THEORY 


OF 


VflTTQQniD     A  DPUfiQ  • 

VUUooUlt[  At[L)illlo , 


BY 


WM,  GAIN,  O.E.,  M,  Am,  Soc,  O.E., 

Professor  of  Mathematics  and  Engineering^  University 
of  North    Carolina. 


SECOND  EDITION,  REVISED  AND  ENLARGED. 


NEW  YOEK: 

D.   VAN  NOSTRAND  CO.,  PUBLISHEES, 
23  MURK  AY  AND  27  WARREN  STREETS. 

1893. 


GENERAL 


COPYRIGHT.  1893. 
D.  VAN  NOSTRAND  COMPANY. 


2  7  6 


TABLE  OF  CONTENTS. 


CHAPTER   I. 

ARTICLE.  PAGE. 

1.  Definitions 11 

2.  Portion  of   arch   taken    for   investiga- 

tion, Hypothesis 14 

8.      Segmeiital  arch  denned 15 

SYMMETRICAL  ARCHES. 

4.  Conditions  of  equilibrium  of  the  arch 

as  a  whole 17 

r>.  Conditions  of  equilibrium  of  the  half 

arch  and  line  of  resistance "2(> 

6.  Formulas  for  thrust  at  the  crown 25 

7.  Scheffler's  method  of   dividing  up  the 

arch  ring  and  spandrel  and  finding 

loads  in  posit.oi:  and  magnitude '27 

s.  Example  of  a  stone  arch  subjected 
only  to  its  own  weight.  Method  of 
tabulating  loads  ;i:id  arms,  and  con- 
structing trial  lin  j  of  resistance 30 

Graphical  method  of  finding  horizon- 
tal thrust. 


II.  TABLE    OF    CONTENTS. 

ARTICLE.  PAGE. 

9.     Same   arch   subjected   to  symmetrical 

live  loads 30 

Method  of  drawing  trial  line  of  resist- 
ance, to  pass  through  any  two  points 
of  half  arch. 

UN-SYMMETRICAL  AECHES. 

10.  Formulas  to  ascertain  position  and 

magnitude  of  thrust  at  crown 40 

Example  1.  Stone  arch  previously  con- 
sidered subjected  to  its  own  weight 
and  an  eccentric  load.  Method  of 
drawing  trial  resistance  line. 

Examples  2,  3  and  4. 

Effects  of  rolling  load  on  piers  in  a 
series  of  arches. 


CHAPTER  1C. 

11.  A  more  correct  method  of  ascertaining 

loads  and  arms 53 

12.  A  more  convenient  method  of  forming 

diagrams  of  forces  and  drawing  a 

trial  line  of  resistance 57 

13.  Usual  method  of  drawing  line  of  resis- 

tance.     Equilibrium  Polygon 64 


TABLE    OF    CONTENTS.  1TI. 

ARTICLE.  PAGE. 

13  (a).  Method  of  constructing  equilibrium 
polygon  when  one  abutment  re- 
action is  given 68 

14.  Special  Properties  of  the  Equilibrium 

Polygon 69 

15.  Example    of    an    arch    subjected    to 

wheel  loads,  by  a  graphical  me- 
thod, to  pass  an  equilibrium  poly- 
gon through  any  three  points.  Pe- 
culiar division  of  arch  ring  and  cor- 
responding computation  of  tables.  78 
Examples. 

CHAPTER  III. 

16.  Properties  of  Curves  of  Resistance 89 

17.  Intersecting  Curves  of  Resistance.. 1)1 

18.  Curve  of  Resistance  corresponding  to 

minimum  and  maximum  horizontal 
thrusts, 92 

19.  Curves   of   minimum    and    maximum 

horizontal    thrusts   for   symmetrical 
arches  with  symmetrical  loads 94 

20.  Curve    corresponding   to    both    mini- 

mum and  maximum   thrust   at   the 
same  time  within  assumed  limits 9(J 

21.  Unit  Stresses  at  any  point  of   a  joint. 
Examples 97 


IV.  TABLE    OF    CONTENTS. 

ABTICLE.  PAGE. 

22.  Three  ways  of  expressing  the  moment 

of  the  thrust  on  any  joint,  about  the 
centre  of  that  joint 104 

23.  Method  of  failure  of  arches. 

Woodbury's  Tables 105 


CHAPTER  IV. 

24.  Historical  note  concerning  authors  who 

have  proposed  to  apply  the  theory  of 

the  solid  arch  to  voussoir  arches 112 

25.  Precise  statement  of  conditions  to  be 

fulfilled  in  order  that  the  theory  of 
the  solid  arch  be  applicable  to  the 
voussoir  arch lir> 

26.  General  table   for  isolated    loads   for 

arches   whose   rise   is   one-fifth   the 

span 117 

General  method  of  procedure. 

27.  Application   to   an    arch   of    100  feet 

span , 1 25 

28.  Table  of  theoretical  depth  of  keystone 

for  stone  arches  whose  rise  is  one- 
fifth  of  the  span 136 

29.  Maximum   intensity   of   unit  pressure 

to  be  allowed  in  stone  or  brick  arches.        143 


TABLE    OF   CONTENTS.  V. 

AEKICLE.  PAGE. 

30.  Formula  for  radius  in  terms  of  span 

and  rise 144 

31.  '  Empirical  formulas  for  depth  of  key 

by  various  authorities 145 


CHAPTER  V. 

32.     Fundamental     equations      concerning 

solid  arches  ' '  fixed  at  the  ends  " 150 

83.     Lemma 154 

34.  Second   general   method  (founded  on 

Eddy's  Constructions  for  the  solid 
arch)  for  testing  the  strength  and 
stability  of  stone  arches.  Live  load.  162 

35.  Example  I.     Arch  of  12.5  ft.  span 164 

36.  Location  of  line  kk 166 

37.  Location  of  line  m  m 166 

38.  Reduction  of  ordinates  of  the  type  m  b .       167 

39.  Proof  that    points  c  satisfy  the    con- 

ditions for  an  "arch  fixed  at  the 
ends" 169 

40.  Location  of   the  Poles  0    and  O    and 

finding    the   centres  of  pressure  on 

the  joints 171 

41 .  Test  of  the  accuracy  of  the  work 1 72 


VI.  TABLE    OF    CONTENTS. 

ARTICLE.  PAGE. 

42.  Example    II.     Arch  of  25  ft.  span 175 

43.  Example  III.     Arch  of  50  ft.  span 176 

44.  Example  IV.  Arch  of  100ft.  span...  177 

45.  Example   V.  Arch  of  100ft.  span  sub- 

jected only  to  its  own  weight.  Inter- 
esting results 177 

46.  No  simple  rule  for  testing  the  stability 

of  arches  approximately 180 

47.  Second  method  not  giving  readily,  the 

the  exact  position   of  live  load  pro- 
ducing the  most  hurtful  effect 182 


APPENDIX. 

Experiments  on  wooden  arches. 184 


PREFACE. 


In  the  present  edition  of  this  work,  the 
method  of  drawing  trial  lines  of  resistance, 
given  in  the  lirst  edition,  is  retained  and 
several  new  chapters  added  containing 
discussions  of  new  constructions,  equilibri- 
um polygons,  properties  of  curves  of  re- 
sistance, unit  stresses,  and  in  the  last  two 
chapters,  two  independent  developments 
of  the  application  of  the  theory  pertaining 
to  solid  arches  "fixed  at  the  ends"  to 
voussoir  arches.  The  Appendix  contains 
the  discussion  of  the  experiments  on  wood- 
en arches,  at  the  limits  of  stability,  given 
in  the  former  edition,  though  here  the 
matter  is  presented  in  a  more  condensed 
form. 

It  will  probably  be  admitted  that  the 
theory*  of  solid  arches,  fixed  at  the  ends, 
applies  directly  to  voussoir  arches,  when 
no  mortar  is  used  in  the  joints  and  the 
voussoirs  fit  perfectly  between  the  skew- 


II. 

backs  when  not  under  stress,  the  resist- 
ance of  the  backing  to  distortion  of  the 
arch  under  stress  being  neglected  and  all 
loads  being  supposed  to  be  transmitted 
vertically  to  the  arch  ring. 

If  the  line  of  resistance  determined  from 
this  theory,  for  the  original  joints,  every- 
where lies  within  the  middle  third  of  the 
arch  ring,  no  further  trial,  on  the  supposi- 
tion of  other  bearing  joints,  has  to  be  made, 
so  as  to  make  the  assumed  and  computed 
joints  agree. 

The  investigation  in  this  treatise  is  lim- 
ited to  this  case. 

The  theory  is  likewise  approximately 
applicable  where  thin  layers  of  cement 
mortar  are  interposed  between  some  or  all 
of  the  arch  stones  that  are  allowed  to 
harden  well  before  the  centres  are  struck. 

For  such  cases,  the  design  of  a  number 
of  arches,  for  spans  from  0  to  160  feet,  for 
a  rise  of  one- fifth  the  span,  are  given  in 
Chapter  IV.  and  the  attention  of  construct- 
ors is  particularly  called  to  this  table  and 
the  comparison  of  results  with  certain  em- 
pirical formulas  as  shown  in  figure  25. 


III. 

It  has  long  been  the  opinion  of  the 
author  that  the  much  heavier  locomotive 
loads  of  to-day  require  greater  depths  of 
keystone  than  are  given  by  many  formulas 
in  current  use,  and  he  submits  that  the 
results  of  Chapters  IV.  and  V.  effectually 
establish  this  position  and  show  the  danger 
of  ignoring  a  theoretical  treatment  of  the 
subject  even  where  only  approximately 
applicable,  as  in  the  case  of  arches  as 
actually  built.  The  loads  may  not  be 
transmitted  vertically  and  the  spandrel 
filling  may  resist  spreading,  and,  in  fact, 
act  partly  as  an  arch ;  but,  it  is  better  not 
to  count  on  these  extra  elements  of  stabil- 
ity, except  as  neutralizing  the  dynamic 
effect  of  the  moving  load  alone,  and  the 
middle  third  limit  prescribed  may  be 
looked  upon  in  the  light  of  introducing  a 
factor  of  safety  against  the  effect  of  the 
loads  regarded  as  static. 

The  author  has  derived  assistance  in  the 
theory  from  Schemer's  Theorie  des  Voutes ; 
also  from  Prof.  Greene's  "Arches,"  and 
Prof.  Eddy's  "New  Constructions  in 
Graphical  Statics. " 


IV. 


He  has  endeavored  to  assign  credit  at 
the  proper  places  in  the  text  to  these  and 
other  authorSo 

CHAPEL  HILL,  May  3d.  l<S(.)o. 


THEORY  OF  VOUSSOIR  ARCHES, 


CHAPTER  I. 

1.  The  voussoir  arch  is  composed  of 
blocks  of  stone,  brick  or  other  material,  in 
the  shape  of  truncated  wedges,  called  arch 
stones  or  voussoirs,  whose  inferior  surface, 
known  as  the  soffit,  is  usually  cylindrical 
and  perpendicular  to  the  radiating  surfaces 
of  contact  called  the  joints.  Between  the 
stones,  mortar,  either  common  or  prefer 
ably  hydraulic,  is  generally  interposed. 

The  lowest  voussoirs  rest  against  the 
abutments  along  the  surfaces  called  the 
skewbacks  or  springing  joints  and  the 
lower  edge  of  these  joints  is  called  the 
springing  line, 

Figure  1  represents  a  section  of  the  arch 
made  by  a  plane  perpendicular  to  a  gene- 
rating element  of  the  cylindrical  surface. 
This  plane  intersects  the  soffit  in  a  curved 
line  called  the  intrados  and  the  exterior 


surface  of  the  arch  ring  in  a  line  called  the 
extrados.  The  intrados  is  generally  an 
arc  of  a  circle  or  two  arcs  of  circles  as  in 
the  gothic  arch,  an  ellipse  or  a  false  ellipse 
(basket-handle)  composed  of  several  arcs 
of  circles  tangent  to  each  other. 

The    crown  of  the    arch    is   the   highest 


part  of  it,  the  crown  joint  is  an  imaginary 
vertical  joint  through  the  crown. 

The  keystone  or  key  is  the  highest  arch 
stone,  extending  generally  either  side  of 
the  crown,  so  that  there  is  no  actual  joint 
at  the  crown. 

The  haunch  or  reins  is  the  part  of  the 
arch  between  the  crown  and  skewback,  and 
the  spandrel  is  the  part  of  the  structure 


13 


above  the  extrados  and  limited  by  the  road- 
way or  other  superior  surface.  The  arch 
is  limited  at  the  ends  or  heads,  by  planes 
called  faces — perpendicular  to  a  generating 
element  of  the  soffit  in  right  cylindrical 
arches,  which  alone  will  be  considered  in 
this  treatise.  The  spandrel  walls  at  the 
faces  are  usually  of  a  superior  quality  of 
masonry,  but  the  space  between  them  is 
often  filled  with  inferior  masonry,  gravel 
or  earth,  called  the  spandrel  filling  or 
simply  the  backing. 

In  large  arches  this  space  is  often 
occupied  by  separate  walls,  running  par- 
allel with  the  roadway  and  connected  by 
flat  stones  or  light  arches,  which  support 
the  material  of  the  roadway. 

The  span  of  the  arch  (Figure  1)  is  the 
perpendicular  distance  between  the  spring- 
ing lines,  and  the  rise  is  the  vertical  dis- 
tance from  the  plane  of  the  springing  lines 
to  the  highest  part  of  the  intrados. 

The  axis  or  axes  of  the  cylindrical  sur- 
face of  the  soffit  will  be  called  likewise  the 
axis  or  axes  of  the  arch. 

When  the  radial  length  of  joint  is  the 


14 

same  throughout,  the  arch  ring  is  said  to 
be  of  uniform  section  and  its  constant 
depth,  measured  radially,  is  the  depth  of 
keystone. 

2.  In  investigating  the  strength  and  sta- 
bility of  right  cylindrical  arches,  it  is  con- 
venient to  consider  a  slice  of  the  arch  and 
load  comprised  between  two  vertical  planes, 
perpendicular  to  the  axis  or  axes  of  the 
arch,  and  one  foot  apart,  and  it  will  be 
understood  that  all  subsequent  figures  refer 
to  such  a  slice  whether  distinctly  stated  or 
not. 

As  the  spandrel  filling  between  the  end 
walls  offers  less  resistance  to  deformation 
of  the  arch  ring  than  the  spandrel  walls  at 
the  faces,  it  is  proper  to  consider  the  slice 
to  be  taken  in  the  interior,  where  the 
backing  is  of  the  least  resisting  character, 
in  order  that  we  may  investigate  the  most 
unfavorable  case. 

As  the  external  forces,  including  the 
weight  of  arch,  will  be  considered  sym- 
metrical with  respect  to  the  vertical  plane 
midway  between  the  faces  of  the  slice,  all 
the  forces  may  be  considered  as  acting  in 


15 


the  medial  plane.  The  same  evidently 
obtains  for  the  internal  stresses  at  the 
joints,  so  that  we  have  simply  to  investi- 
gate the  equilibrium  of  forces  in  one 
plane. 

The  hypothesis  will  be  made  that  the 
weight  per  cubic  foot  of  the  arch  stones  is 
the  same  throughout  ;  similarly  for  the 
spandrel  filling.  This  is  not  exactly  true, 
but  is  certainly  near  enough,  when  stones 
are  selected  from  the  same  quarry,  espec- 
ially when  the  arch  is  investigated  as  to  the 
action  of  the  very  heavy  eccentric  loads  of 
to-day ;  for  any  irregularity  can  be  sup- 
posed included  in  the  hypothetical  loading 
(never  exactly  realized)  with  perfect 
safety. 

Other  approximations  will  be  carefully 
noted  as  we  proceed. 

o.  In  constructing  voussoir  arches  a 
wooden  frame  or  centre  is  built  between 
the  abutments,  whose  upper  convex  surface 
exactly  coincides  with  the  soffit  of  the  arch 
to  be  built.  The  lower  arch  stones  are 
laid  first,  and  it  is  a  matter  of  experience 
that  in  a  semi-circular  arch  the  arch 


16 

stones  can  be  laid  successively  up  to  about 
half  the  rise  without  any  centering  what- 
soever. This  corresponds  to  the  joint 
which  makes  an  angle  of  30°  with  the 
horizontal,  or  00°  with  the  vertical.  The 
central  portion  of  12 i.°  has  to  be  supported 
by  the  centre,  until  finally  the  keystone 
(which  should  exactly  fit  the  remaining 
space)  is  driven  in  and  the  centre  removed. 

If  the  spandrel  wall  is  built  up  to  the  3<  ° 
joint  as  solidly  as  the  abutment,  as  should 
always  be  done,  the  arch  and  spandrel  be- 
low this  joint  can  be  considered  as  rigid 
and  a  part  of  the  abutments;  hence  the 
true  arch  is  the  central  portion  of  12i.° 
total  amplitude. 

On  this  account  it  is  convenient  to  de- 
fine a  segmented  arch  as  one  whose  total 
amplitude,  or  angle  between  the  planes  of 
its  extreme  joints,  is  equal  to  or  less  than 
120°. 

The  pressure  of  one  arch  stone  against 
another  at  the  joints  gives  rise  to  molecular 
stresses,  represented  by  the  little  arrows  in 
figure  1.  It  is  our  object  first  to  find  the 
resultants  of  these  stresses  on  each  joint. 


17 

The  thrust  at  the  crown  is  thus  the  result- 
ant of  the  stresses  on  one  side  of  the 
crown  joint  exerted  against  and  balanced 
by  the  stresses  on  the  other  side  of  the 
joint.  Similarly  for  any  other  joints. 

SYMMETRICAL  ARCHES. 

4.  We  shall  consider  first  an  arch  formed 
of  two  branches  AC,  130  (Figure  2), 
symmetrical  and  placed  in  juxtaposition, 
and  comprised  between  two  parallel  ver- 
tical planes  perpendicular  to  the  axis  of 
the  arch,  the  arch  being  right  cylindrical. 
This  arch,  composed  of  voussoirs  in  the 
shape  of  wedges,  leans  against  two  abut- 
ments at  its  extremities  A  and  B,  and  is 
loaded  not  only  with  its  own  weight  but 
with  any  other  weight  whatsoever,  distrib- 
uted symmetrically  on  either  side  of  the 
crown  C.  The  mass  of  the  arch  is  sub- 
ject to  the  laws  of  friction  in  its  joints. 
The  adherence  of  the  mortar,  interposed 
between  the  voussoirs,  being  difficult  to 
estimate  will  not  be  considered.  As  the 
two  half  arches  are  symmetrical  as  to  the 
crown  C,  it  is  clear  that  the  points  of 


18 


application,  A  and  B  of  the  reactions  Ra 
and  R2  of  the  surfaces  of  support,  will  be 
also  symmetrical  in  relation  to  the  vertical 


passing  through  the  crown,  and  that  the 
line  AB  will  be  horizontal,  in  whatsoever 
manner  the  points  A  and  B  may  vary 
upon  the  surfaces  of  support. 


Now  the  weight  of  arch  and  load  =  P, 
acting  downwards  through  the  crown  C, 
together  with  the  reactions  Rr  and  R2 
acting  upwards,  form  a  system  of  forces 
in  equilibrium,  and  because  Hl  and  R2 
must  meet  P  in  a  point  D  they  are 
equally  inclined  to  P  and  hence  are  equal. 

If  we  decompose  the  reactions,  R:  and 
R2,  into  their  horizontal  and  vertical  com- 
ponents Pj,  Q15  P2,  Q2,  we  should  have 
P1  =  P2=  the  weight  of  half  arch  with  its 
load,  and  the  thrust  QjZuQ,,. 

Let  us  consider  now  one  of  the  two 
halves,  for  example  AC.  Let  EPj  be  the 
vertical  passing  through  the  centre  of 
gravity  of  this  half  with  its  load;  to  hold 
this  mass  in  equilibrium  it  is  necessary 
that  there  exists  at  the  crown  a  force 
whose  direction  CE  passes  through  the 
point  of  intersection  E  of  the  vertical  EP, 
with  the  direction  of  the  reaction  Rr 

As  the  vertical  component  of  R:  equals 
the  weight  Pl  acting  through  E  and  since 
by  the  laws  of  statics,  the  algebraic  sum 
of  the  horizontal  components  of  the  forces 
acting  on  the  half  arch  AC,  must  equal 


zero,  the  thrust  at  the  crown  C  of  the 
arch  is  necessarily  equal  to  the  second 
component  Q1?  of  the  reaction  R.  and  must 
be  horizontal  as  it  is. 

From  what  preceeds  we  are  allowed  to 
consider  only  a  half-arch,  leant  against  a 
FIG.  3. 


fixed  surface  at  A,  and  solicited  by  a  hori- 
zontal force  at  C. 

5.  (Figure  3.)  Let  ab,  oj)^  «2/>2,  be  the 
joints  of  an  arch;  P19  P.,  the  vertical 
directions  of  the  weights  of  the  parts  ab 
b^a,\  <fbb2  <?2;  including  the  loads  on  the 


parts  bbl9  bb^ ;  P1?  P,  acting  through 

their  common  centres  of  gravity. 

The  horizontal  force  Q,  at  the  crown, 
combined  with  the  reaction  R2  at  the  j  oint 
«8&2  holds  the  part  abb9a9  in  equilibrium 
and  similarly  for  the  reactions  on  other 
joints. 

At  the  points  where  the  direction  of  Q 
cuts  P,,  P2,  combine  those  forces  with  Q 
as  shown  in  the  figure;  the  resultant  of  Q 
and  Pj  cuts  joints  alb1  at  Aj,  which  is 
therefore  the  centre  of  pressure  on  that 
joint.  As  the  weight  abb2az  with  its  load 
equals  P2  and  is  the  weight  on  joint  a2  &2, 
the  resultant  of  P2  and  Q  will  give  the 
force  acting  on  ajb^  in  direction,  position 
and  magnitude ;  it  cuts  «262  at  A2,  which 
is  therefore  the  centre  of  pressure  on  that 
joint. 

In  the  same  way  the  resultants  and  cen- 
tres of  pressure  on  all  the  joints  may  be  de- 
termined. A  broken  line  connecting  these 
centres  of  pressure  on  the  various  joints 
will  be  called  the  line  of  the  centres  of 
pressure  or  more  briefly,  the  line  of  re- 
sistanee. 


For  voussoirs  indefinitely  small,  it  ap- 
proaches indefinately  a  curved  line  and 
will  be  called  the  curve  of  the  centres  of 
pressure  or  curve  of  resistance. 

That  granted,  in  order  that  the  arch 
may  remain  in  equilibrium,  it  is  necessary : 

(1.)  That  the  points  of  intersection 
C',  A,,  A2,  fall  in  the  interior  of  the 
respective  joints  ab,  a}bl9  ajby  If  for  any 
joint  this  is  not  so;  e.  </.,  if  the  point  A] 
was  above  bl9  the  mass  abb^  al  would  then 
turn  around  the  edge  61?  as  an  unresisted 
couple  would  be  formed.  To  explain  : 
suppose  the  resultant  Hl  to  pass  outside 
of  joint  albly  conceive  two  equal  opposed 
forces,  each  equal  to  R,  to  act  at  edge  b,- 
this  does  not  disturb  the  equilibrium; 
then  R,  the  force  acting  through  A, 
(which  is  outside  the  joint)  with  its  equal 
but  not  directly  opposed  force  at  bl9  would 
form  the  unresisted  couple  in  question 
which  causes  overturning. 

(2.)  That  the  directions  c^A^  c2A2  of 
the  pressures  upon  the  joints  do  not  make 
angles  with  the  normals  to  the  respective 
joints  which  exceed  the  angle  of  friction. 


If  it  was  not  so,  sliding  at  the  joints  in 
question  would  occur  of  the  mass  above  or 
below. 

However,  the  friction  of  the  materials 
usually  employed  in  construction  is  suffi- 
ciently great  to  not  give  cause  for  fear 
as  regards  sliding,  generally. 

It  is  very  easy  to  alter  the  direction  of 
the  joints  should  sliding  be  apprehended, 
hence  it  will  not  be  considered  further. 

The  two  former  requirements  refer  to 
stability^  supposing  no  crushing  of  the 
material  occurs.  The  last  requirement 
pertains  to  the  strength  of  the  masonry 
at  a  mortar  joint. 

(3.)  The  resultant  of  the  pressure  on 
any  joint  must  not  pass  so  near  the  edge 
that  crushing  of  the  mortar  or  the  stone 
(or  brick)  may  occur.  Of  course  in  prac- 
tice a  certain  factor  of  safety  would  be 
used,  but  as  this  subject  will  be  treated  in 
full  later,  we  shall  only  remark  here  that 
reasons  will  be  given  further  on  why  the 
true  line  of  resistance  should  not  depart 
at  any  joint  more  than  one- sixth  the  depth 
of  joint  from  its  centre,  or  in  other  words, 


24 

should  everywhere  be  confined  to  the  mid- 
dle third  of  the  arch  ring  and  in  fact  it  is 
best  to  use  still  narrower  limits. 

It  is  to  be  remarked  that  the  foregoing 
theory  does  not  require  horizontal  resist- 
ance in  the  spandrel,  which  is  not  gener- 
ally built  with  the  same  care  that  is  taken 
in  the  construction  of  the  arch  stones,  and 
thus  cannot  generally  be  regarded  as  un- 
yielding; hence  when  a  line  of  resistance 
such  as  C'Aj  A.,  passes,  somewhere,  out  of 
the  arch  ring,  a  serious  derangement  of 
the  arch  may  occur,  even  though  the 
spandrel  may -prevent  its  falling:  hence  it 
appears  to  be  a  poor  construction  to  build 
such  an  arch  in  preference  to  an  arch  in 
which  the  resultant  pressures  on  the  joints 
everywhere  keep  within  the  limits  pre- 
scribed. This  will  be  adverted  to  again. 

If  no  line  of  resistance  can  be  dniwn 
within  the  prescribed  limits,  or  crushing  is 
feared  on  any  joint,  the  depth  of  the  vous- 
soirs  must  be  increased  on  part  or  the 
whole  of  the  arch;  or  the  profile  may  be 
altered;  or,  finally,  we  may  combine  both 
of  these  methods  to  secure  stability. 


6.  We  shall  now  enter  into  detail  to 
show  how  lines  of  resistance  can  be  drawn 
through  various  points  of  the  arch  ring. 

Let  us  consider  as  in  Art.  2  a  slice  of  the 
arch  1   unit  thick.     In  Fig.  4,  let  Q=the  , 
horizontal   thrust   at   the   point  M  of  the ! 


crown  joint  (compare  Fig.  3  throughout) ; 
</,  its  vertical  distance  below  the  apex  C; 
P= weight  of  arch  and  load  C  A  on  joint 
at  A;  a — horizontal  distance  from  A,  tlK- 
centre  of  pressure  on  the  joint,  to  the  ver- 
tical passing  through  the  centre  of  gravity 
of  P;  h— vertical  distance  between  C  and 


A;  b  =  vertical  distance  between  M,  the 
point  of  application  of  Q  at  the  crown, 
and  A. 

If  we  consider  another  point  A'  of  the 
mrve  of  resistance,  at  a  vertical  distance 
above  A=e,  we  shall  have  an  analogous 
notation  P',  a',  b'. 

If  we  know  the  points  M  and  A,  we  have, 
taking  moments  about  A, 

«P  =  bQ    .-.  Q  =         ............  (1). 


If   we    know   any   two  points  as  A,  A', 
=bQ  =  (£'-f  e\Qu  also  a'  P  =  b'  Q, 

~ 


hence,  Q  =  —  ?~  .............  (2). 


Having  obtained  from  eq.  (1)  or  eqs.  (2) 
and  (o),  Q  and  its  point  of  application  at 
the  crown,  we  find  where  the  resultant 
pressures  cut  each  joint  as  in  Art.  3. 

If  the  first  curve  drawn  passes  outside 
of  the  prescribed  limits  in  one  or  more 
places,  take  points  on  the  limiting  curves 


27 

opposite  the  points  of  maximum  departure, 
and  by  eqs.  (^)  and  (3),  pass  a  curve 
through  two  of  these  points. 

If  the.arch  is  stable  at  all,  it  will  almost 
invariably  be  found  in  practice  that  the 
last  curve  so  drawn  will  fulfill  the  required 
conditions  of  remaining  in  the  prescribed 
limits.  If  not  a  third  approximation  may 
be  tried,  and  so  on. 

This  is  Dr.  Scheffler's  method  of  draw- 
ing a  line  of  resistance  within  prescribed 
limits  and  in  practice,  after  becoming 
familiar  with  the  leading  cases,  the  first 
trial  is  generally  sufficient. 

7.  Let  us  proceed  to  show  how  to  find 
the  centres  of  gravity  of  the  weights  abb^ 
d^  abb^a^  (Fig.  3),  as  also  the  magnitudes 
of  those  weights.  If  the  arch  is  loaded 
with  any  weights,  reduce  them  to  the 
same  specific  gravity  as  that  of  the  masonry 
of  the  bridge  supposed  homogeneous,  as 
follows :  Lay  down  these  weights  in  their 
exact  positions  on  the  arch,  and  alter  the 
vertical  dimensions  to  conform  to  the 
specific  gravity  of  the  stone.  We  shall 
thus  substitute  blocks  of  stone,  by  scale, 


28 

for  the  surcharge  of  earth,  water,  etc.3  or 
the  rolling  load. 

We  now  divide  the  horizontal  through 
A  (Fig.  5)  into  an  appropriate  number  of 
parts  and  through  these  points  of  division 


draw  verticals  from  the  intrados  to  the 
curve  DE  that  limits  above  the  reduced 
load. 

Regard  each  trapezoid  DGG'D'  as  a 
rectangle,  and  calculate  its  surface  by 
multiplying  its  horizontal  width  AA'  by 


29 

the  mean  vertical  dg.  Next  regarding  the 
centre  of  gravity  of  each  trapezoid  as  that 
of  the  corresponding  rectangle,  we  shall 
find  the  centre  of  gravity  of  the  trapezoid 
DD'G'G,  for  .example,  to  be  upon  the 
mean  vertical  dg,  which  equally  divides 
the  horizontal  AA'.  Draw  through  C,  the 
joint  CH;  the  weight  DD'G'G  will  be 
considered  as  resting  on  the  joint  CH, 
which  is  in  excess  by  the  small  triangle 
CG'H,  an  error  too  small  to  be  regarded 
in  flat  arches. 

Scheffler  gives  a  graphical  construction  for  correcting 
the  joints  to  correspond  better  to  the  weights,  but  it  is  very 
tedious  in  practice  and  defaces  the  drawing  with  too  many 
many  lines,  so  it  is  not  given.  In  article  11  will  be  found  a 
method  that  will  insure  all  desirable  accuracy  for  any  form 
of  arch. 

The  assumption  that  the  whole  weight 
vertically  over  a  voussoir  acts  on  it  is  not 
strictly  true,  whether  the  spandrel  is  of 
earth  or  masonry ;  for  part  of  it  is  doubt- 
less distributed  in  a  different  manner  by 
aid  of  the  friction  of  the  particles,  or  in 
some  cases  from  the  spandrel  itself  acting 
as  a  sort  of  arch.  Again,  any  horizontal 
spreading  of  the  haunches  of  the  arch  is 


30 


partially  hindered  by  horizontal  resistances 
in  the  spandrels.  These  influences  though 
are  on  the  side  of  safety,  particularly  when 
the  dynamic  effects  of  moving  loads  are 
considered.  In  the  case  of  culverts  the 
earth  exerts  a  thrust,  as  to  which  and  the 
treatment  of  culverts  and  tunnel  arches, 
see  Science  Series  No.  42. 

Assuming  the  loads  over  the  arch  to  act 
as  above,  we  multiply  the  surface  of  each 
trapezoid  by  the  horizontal  distance  of  its 
centre  of  gravity  from  A ;  the  sum  of  these 
moments  divided  by  the  sum  of  the  trape- 
zoid surfaces  (which  are  also  the  volumes), 
will  give  the  horizontal  distance  from  A 
to  the  centre  of  gravity  of  the  whole  part 
considered.  This  method  will  thus  give  us 

the  weights  P1?  P2, (Fig.  3),  as  well  as 

the  horizontal  distance  of  their  centres  of 
gravity  from  the  crown. 

8.  First  Example. — Let  us  illustrate  by 
an  example  of  a  railroad  bridge  (Fig.  1.5)  of 
50  ft.  span  and  10  ft.  rise,  the  arch  being 
a  segment  of  a  circle ;  voussoirs  2.5  ft.  deep ; 
the  spandrel  walls,  of  the  same  specific 
gravity  as  the  voussoirs,  rising  2. S3  ft. 


31 


above  the   summit  of  the  arch  rin5.      i'he 


arch  is  7  ft.  thick,  but  we  shall  consider 
but  a  vertical  slice  of  it  1  ft,  thick. 


32 

In  the  following  table  the  first  column 
gives  the  number  of  the  joint  from  the 
crown;  the  second  (w)  the  width  of  the 
horizontal  divisions  AA',  A'A"-  -  of  Fig.  >"> ; 
the  third  (v)  the  corresponding  mean 
heights  dg  -  -  -;  the  fourth  (s),  the  prod- 
uct of  these  dimensions,  giving  thus  the 
surface  of  each  trapezoid.  Column  (c) 
gives  the  distance  of  the  centre  of  grav- 
ity of  each  trapezoid  from  the  crown ; 
column  (m)  the  product  of  (s)  and  (c). 
Now  we  cumulate,  going  from  the  crown, 
in  the  next  two  columns,  these  surfaces  (s) 
and  products  (s  X  c) ;  column  (S)  being 
formed  by  adding  the  surface  of  each 
trapezoid  to  the  total  surface,  just  found, 
which  precedes  it.  The  last  quantity  in 
column  (S)  should  =  sum  of  col  urn  (s). 

In  the  same  way  column  (M)  contains 
the  continued  sum  of  column  (m),  and 
hence  its  last  number  should  equal  the 
sum  of  column  (m).  Dividing  now  the 
numbers  in  column  (M)  by  the  corres- 
ponding ones  in  column  (S)  we  get,  col- 
umn (C),  the  horizontal  distances  of  the 
centre  of  gravity  of  each  weight  Pl5  P2 — 


— ,  corresponding  to  joints  1,  2  -  -  -,  from 
the  crown. 


1 

w 

v 

s 

C 

m 

S 

M 

C 

1 

2 
3 
4 
5 

G 

5 

5 
5 
5 
5 
1.75 

5.4 
6.1 
76 
9.8 
13.2 
14.5 

27. 
30.5 
38. 
49. 
66. 
25.4 

2.5 
7.5 
12.5 
17.5 
22.5 
25.9 

67.50 

228.75 
475. 
857.50 
148  5. 
657.86 

27. 
57.5 
95.5 
144.5 
210.5 
235.9 

67.50 
296.25 
771.25 
1628.75 
8113.75 
3771.61 

2.5 
5.1 
8  1 
11.  « 
14.7 
16. 

235.9 

3771.61 

The  preceding  table  shows  that  the 
surface  (or  volume,  for  a  slice  1  ft.  thick) 
of  the  half  arch  with  its  load  equals  235.(.i 
sq.  ft.;  its  moment  as  to  the  crown  is 
3TTUJ1  and  the  distance  of  its  centre  of 
gravity  from  the  joint  at  the  crown  is  10  ft. 
Let  it  be  required  to  pass  a  curve  of  resist- 
ance through  the  crown  joint,  K  of  its 
depth  from  the  summit  of  the  arch  and 
through  joint  (>  at  }-3  of  its  depth  above  its 
lowest  point.  By  measurement  on  the 
drawing  (Fig.  9)  we  finda=25.i> — 16=1). (>, 
6=11.2.  We  have  also  P=235.9  cubic 
feet  of  stone;  hence  by  formula  (1),  Art.  0, 


235.9x96 
~lT2~ 


=  202   cubic  ft.  stone 


which  may  be  reduced  to  tons,  when  de- 
sired, by  multiplying  by  the  weight  in 
tons  of  a  cubic  foot  of  stone. 

If  now  at  the  points  of  intersection  of 
the  horizontal  through  the  point  of  appli- 
cation of  Q  at  the  crown,  with  the  verti- 
cals passing  through  the  centres  of  grav- 
ity of  the  surfaces  given  in  column  (S), 
[Pj,  P2,  -  -  -,  of  Fig.  3],  we  combine  these 
weights  with  Q,  the  points  of  intersection 
of  the  resultants  of  Q  with  these  weights 

P,,  Pjj, ,with  the  corresponding  joints, 

will  be  points  in  the  curve  of  resistance 
sought. 

For  example,  to  determine  where  the 
line  of  resistance  cuts  joint  4,  lay  off  the 
distance  in  column  (C),  11.3  horizontally 
from  the  crown,  then  on  a  vertical  lay  off 
upward  from  this  point  the  corresponding 
weight  on  joint  4,  given  in  column  (S) 
144.5 ;  drawing  a  horizontal  line  through 
the  last  point  found=Q  =  20^,  we  get  the 
resultant  by  completing  the  triangle  of 
forces. 

Producing  this  resultant  to  intersection 
with  joint  4,  will  give  the  centre  of  pres- 


sure  on  that  joint.  It  will  be  advisable, 
in  practice,  to  prick  off  the  centres  of 
gravity,  taken  from  column  (C),  at  one 
operation  and  number  each  one  with  the 
number  of  the  corresponding  joint  to 
avoid  mistake. 

On  continuing  this  construction  for 
each  joint,  we  shall  find  that  the  line  of 
resistance  remains  within  the  inner  third 
of  the  arch  ring. 

It  may  be  remarked  that  the  small  tri- 
angle mentioned  is  in  excess  only  for  the 
joint  in  question;  thus  this  error  is  not 
carried  on. 

The  ordinary  method  of  constructing  a 
line  of  pressures  is  to  combine  any  result- 
ant with  the  next  weight  following,  re- 
garded as  concentrated  at  its  centre  of 
gravity.  By  this  construction  any  small 
error  in  draughting  is  carried  on,  whereas, 
by  the  former  method,  it  is  confined  only 
to  the  joint  where  it  occurs  first. 

With  accurate  instruments  and  care, 
using  a  sufficiently  large  scale,  this  meth- 
od should  answer  all  the  requirements  of 
accuracy,  and  will  generally  be  found  the 


shortest  in  the  end;  whereas,  with  many 
joints,  it  is  difficult  to  locate  this  curve 
precisely  by  the  ordinary  method. 

We  have  made  use  in  the  above  example, 
of  formula  (1)  to  compute  the  thrust  at 
the  crown.  This  can  preferably  be  found 
without  computation,  as  follows:  at  the 
point  H  on  the  horizontal  through  the 
crown,  lay  off,  to  the  scale  of  force,  verti- 
cally upwards  the  weight  of  half  arch  and 
load  =  #35.!.)  and  through  the  extremity  of 
the  line,  draw  an  indefinite  horizontal. 
The  intersection  of  the  latter  with  the  line 
drawn  through  the  point  <>  first  mentioned 
and  the  assumed  centre  of  pressure  on 
joint  f>,  will  cut  off  a  distance  on  this 
horizontal,  equal  to  Q  to  the  scale  of  force. 
This  value  of  Q  is  then  to  be  laid  off  in 
constructing  all  the  other  triangles  of 
force.  This  method  can  likewise  be  fol- 
lowed in  the  next  example  if  preferred. 

9.  Second  Example. — (Fig.  7.)  Suppose  a  load  of  two 
40-ton  engines,  one  on  each  side  of  the  crown,  over  divis- 
ions 2,  3,  and  4,  i.  e.,  15  ft.  along  the  rails.  We  shall  sup- 
pose it  to  bear  only  on  6  ft.  of  the  thickness  of  the  viaduct. 
Calling  the  weight  of  a  cubic  foot  of  stone  =  .07  ton  and  A, 
the  height  of  the  block  of  stone  15  ft.  long  by  6  ft.  wide  that 
is  required  to  weigh  as  much  as  one  engine;  we  have 
6  X  15  X  h  .07  =40  .-./*=  6.3. 


We  now  form  the  following  table  which  refers  to  Fig.  7, 
which,  as  the  arch  and  load  is  symmetrical,  represents,  as 
before,  only  one-half  the  arch. 


Joint 

w 

v 

s 

C 

in 

S 

M 

C 

1 
2 
3 
4 
5 
6 

5 

5 
5 
5 
5 
1.75 

5.4 

12.4 
14. 
16. 
13.2 
14.5 

27 
62 
70 
80 
66 
25 

2.5 
7.5 
12.6 
17.5 
22,5 
25.9 

67 
468 

869 

HI  18 
1485 
658 

•21 

159 
239 

:•;•  >5 
330 

67 
532 
1401 

2809 
4294 
4952 

2.5 
6. 

11.8 

14.1 

15. 

330 

4952 

A  line  of  resistance  passing  through  the  middle  01  the 
crown,  the  point  on  the  springing  joint,  as  before,  will  be 
found  to  be  contained  inside  of  the  limiting  curves,  and  is 
drawn  as  in  Fig.  7,  taking  care  to  lay  off  the  centres  of  gravity 
on  the  prolongation  of  Q.  We  find  in  this  case  a  =  25. 6  —  15 
=  10.6,  P=330,  6=10.7. 

.Q  =  880  X  10.6  =327=c=23tone. 

If  it  is  desired  to  draw  the  curve  corresponding  to  the 
minimum  of  the  thrust  in  the  limits  chosen  (see  Art.  19.) 
we  resort  to  equations  (2;  and  (3).  As  the  nearest  approach 
of  the  last  line  of  pressures  drawn  to  the  outside  limiting 
curve,  is  at  joint  2  ;  pass  a  curve  of  resistance  through  the 
point  of  intersection  of  that  outside  limiting  curve  with  the 
second  joint  and  the  previous  point  at  the  springing  joint. 
We  find  P=  330,  a—  10  6,  <?=9.8  and  from  table  2,  column 
(S)Pi  =  S9;  from  column  (c)  and  the  drawing  a!  =9. 8  —  6 
=  3.8. 

From  (2),  Art.  6, 

aP_aiPl       3498-338 


From  (3) 


=11.93  — 


9.8 
3498 


Laying  off  this  latter  distance,  from  the  summit  of  the 
arch  ring,  downwards,  we  draw  the  curve  as  before.  It  is 
everywhere  within  the  proper  limits,  and  of  course  touches 
the  upper  middle  third  limit  at  joint  2  and  the  lower 
middle  third  limit  at  joint  6,  as  assumed. 

If  we  suppose  a  weight  of  13.3  tons  to  rest  on  division  3 
on  both  sides  of  the  crown,  along  5  ft.  of  the  length  of  the 
rails,  we  shall  find  by  forming  a  table  and  constructing  the 
line  of  resistance  as  in  the  last  case  above,  that  it  passes 
slightly  below  the  upper  limit  at  the  crown,  and  is  every- 
where contained  in  the  middle  third  of  the  arch  ring. 

A  curve  of  resistance  for  a  uniform  load  of  1.5  tons  per 
foot  along  the  whole  length  of  the  bridge  can  be  drawn  to 
follow  very  closely  the  curve  drawn  in  the  first  example. 

One  or  two  more  suppositions  of  isolated  weights,  sym- 
metrically placed,  were  made,  but  in  all  cases  it  was  found 
that  a  curve  of  resistance  could  easily  be  drawn  in  the  inner 
third  of  the  arch  ring.  The  thrust  is  too  small  to  fear  crush- 
ing, and  the  directions  of  the  thrust  are  inclined  to  the  nor- 
mals of  the  arch  joints  at  angles  much  smaller  than  the 
angles  of  friction,  hence  sliding  is  not  to  be  feared. 

The  curves  of  resistance  drawn  in  the 
preceding  examples  are  not  necessar- 
ily the  true  ones,  otherwise  we  should 
at  once  conclude  that  thus  far  the  arch 
had  stability.  The  true  curve  depends 
upon  the  elastic  yielding  of  the  arch  to  the 
weights  acting  on  it  and  we  shall  see  later 
how  the  aid  of  a  few  principles  from  the 
theory  of  elasticity  will  enable  us  to  locate 
it  approximately.  For  the  present,  we 
shall  continue  the  subject  by  showing  how 


any  trial  line  of  resistance  can  be  made  to 
pass  through  any  three  points  of  an  arch 
ring,  either  unsymmetrical  or  unsyminet- 
rically  loaded.  The  method  first  given 
below  requires  the  solution  of  some  equa- 
tions: subsequently  in  Art.  15  a  purely 
graphical  method  is  developed  which  will 
doubtless  generally  be  preferred. 

UNSYMMETRICAL  ARCHES. 

10.  In  unsymmetrical  arches,  or  arches 
unsymmetrically  loaded,  there  is  a  joint, 
EF  (Fig.  8),  generally  near  the  crown,  at 
which  the  thrust  is  horizontal.  Where 
the  arch  is  only  solicited  by  vertical  forces, 
by  compounding  them  with  this  thrust,  as 
before,  we  find  the  resultants  on  every 
joint,  and  it  is  evident  in  this  case  that  the 
horizontal  thrust  is  the  same  all  through 
the  arch. 

It  is  more  convenient  however,  to  find 
the  inclined  thrust  at  the  crown  and  com- 
bine the  partial  weights  with  it,  to  find 
the  resultant  on  each  joint. 

Problem :  To  find  this  inclined  thrust 
and  its  point  of  application : 


Let    I,   L,    K  be   three    points   through 

which  to  pass  a  curve  of  resistance  (Fig. 

(8),  ACB  a  horizontal  line  drawn  through 

the  highest  point  of  the  extrados,  and  let 

•  there  be: 

Q,  the  horizontal  thrust,  i.  e.  the  horizon- 
tal component  of  any  one  whatsoever  of 
the  pressures  acting  through  I,  L,  K. ; 

P,  the  vertical  component  of  the  pres- 
sure S  at  the  crown  joint,  which  will  be 
considered  positive,  if  it  is  directed  up- 
wards, as  regards  pressure  from  the  right 
part  upon  the  left. 

Let  <?,  be  the  vertical  distance  of  the 
point  of  application  of  S  below  the  hori- 
zontal ACB;  (fv  hi'y  9v  h* ;  9v  h*>  the  hori- 
zontal and  vertical  co-ordinates  of  I,  K 
and  L  as  to  the  point  C  as  the  origin ; 

P1,  P11,  P111,  the  vertical  components  of 
the  pressures  acting  through  I,  K  and  L; 
P1?  P2,  P3,  the  weights  of  the  segments  C 
,  I,  CK,  CL,  with  their  loads ; 

#i>  <32>  as,  the  horizontal  distances  of 
the  centres  of  gravity  of  these  segments 
CI,  CK,  CL,  from  the  points  I,  K  and  L 
respectively. 


43 

To  abbreviate,  let  us  put: 
h1-q=b1  ha-q=b2 


Observe  that  arch  CI  is  in  equilibrium 
under  the  action  of  the  left  reaction  tPl 
acting  up  and  Q  acting  to  the  right  being 
its  components)  the  thrust  S  at  the  crown 
(P  acting  up  and  Q  acting  to  left  being  its 
components)  and  the  weight  of  arch  CI 
and  load  Pl  acting  downwards. 

Similarly,  the  part  CK  is  in  equilibrium 
under  the  action  of  P''  (acting  up)  and  Q 
(acting  to  left)  at  K,  the  force  P  (acting 
down)  and  Q  (acting  to  the  right)  at  the 
crown  and  the  weight  P2. 

Lastly,  the  part  CL  can  be  dissociated 
from  the  rest  and  conceived  to  be  in 
equilibrium  under  the  action  of  the  reac- 
tion at  L  acting  upwards,  the  forces  P 
(acting  up)  and  Q  (acting  to  the  left)  at 
the  crown  and  the  weight  P3  of  CL  and  load. 

Balancing  vertical  components  for  the 
parts  CI  and  CK  respectively,  we  have, 
P'-fP^P,  ........................  (1) 

P"_P  =  Pt  .......................  (2) 


44 

Next  taking  moments  about  I,  K  and  L 
of  the  forces  holding  in  equilibrium  the 
part  CI,  CK  and  CL  respectively. 


a,P8+g8P=b8Q  .................  .  ...........  (4) 

a3  P3-g3  P=b3  Q  ............................  (5) 

If  the  third  given  point  L  of  the  curve 
of  pressures  is  upon  the  joint  at  the  crown 
C,  the  value  of  q  is  known,  and  we  have  : 
gs  =  o,  h3  =  q,  P8  =  O.  From  eqs.  (3)  and 
(4)  we  find 


If  L  is  not  upon  the  joint  at  the  summit, 
we  find  * 


T> 


Q 


.(10) 


*  Add-(4)  and  (3)  and  call  the  sum  eq.  (11);  also  subtract 
(5)  from  (3.)  Place  the  values  of  Q  equal  to  each  other  in  this 
last  eq.  and  eq.  (11);  reducing,  bearing  in  mind  that  do-c?:j 
=dlf  e.2-e3  =  ci,  &c.,  we  find  P  as  in  eq.  (8).  Substitute 
this  value  of  P  just  found  in  eq.  (11)  and  deduce  Q.  which 
gives  eq.  (9).  Eqs.  (6)  and  (5)  are  only  particular  cases  of 
eqs.  (8),  (9)  and  (10)  when  P;j  =  O. 


Example  1.  Fig.  9  represents  the  same 
viaduct,  before  considered  in  Art.  8,  with 
a  load  of  40  tons  on  15  feet  of  length  over 
divisions  "<?,  3  and  4,  on  one  side  of  the 
arch  only.  The  table  of  Art.  8  refers  to  the 
right  half  of  the  arch :  the  table  of  article 
0  to  the  left  side. 

Let  us  pass  a  curve  of  resistance  through 
the  middle  of  the  crown  and  through  a 
point  on  each  springing  joint,  y^  depth 
joint  above  its  lower  edge. 

We  find  from  the  drawing  and  tables. 

9^=9%-^^  £1  =  £2:=10.75 

P]  =  330,  A  =  15,  a,  =  10.6 

Pi  =  ^3(>,  p2  =  l<>,  as=   9.6 

From  ((>) : 

p=«,^q3*,P>=24  cubic  ft<  of  gto 
ffA-r^ffi 

from  (T): 

Q— ^£l-^i?==^68  cubic  ft.  of  stone; 

From  M,  the  middle  of  the  crown  joint, 
lay  off  downwards  MN=P,  alsoNH=Q  on 
the  horizontal  through  N;  MH  will  then 
represent  tl  on  the  crown  joint 


47 

in  direction,  position  and  magnitude;  and 
by  combining  it  with  the  weight  of  each 
artificial  voussoir  and  load,  on  each  side  of 
the  crown,  each  acting  through  its  centre 
of  gravity,  we  evidently  obtain  the  result- 
ants on  the  various  joints  in  direction,  posi- 
tion and  magnitude,  and  therefore  can 
trace  the  curve  of  pressures.  For  example, 
to  find  the  resultant  on  the  third  joint  on 
the  left  side  of  the  arch :  draw  a  horizon- 
tal line  through  M  and  lay  off  on  it  the 
distance  of  the  centre  of  gravity  of  the 
three  first  divisions,  from  M,  which  by 
Table  2  ( Art.8) ,  column  C,  is  found  to  be  8.8. 

Draw  a  vertical  through  this  point  and 
from  its  point  of  intersection  with  MH,  lay 
off  upwards  the  weight  159  (column  S)  of 
the  three  divisions  in  question. 

From  the  upper  extremity  of  this  last 
line  draw  a  line  ||  and  equal  to  MH;  com- 
pleting the  parallelogram  of  forces  as  per 
figure,  the  point  where  the  resultant  cuts 
joint  3  is  the  centre  of  pressure  of  that 
joint,  and  the  resultant  is  given  in  magni- 
tude, position  and  direction  by  the  diago- 
nal. 


48., 

The  construction  for  the  other  joints  is 
the  same. 

The  nearest  approach  of  the  curve  of 
pressures  to  the  extraclos  is  on  joint  1,  of 
the  left  side  of  the  arch,  where  it  is  only 
three-tenths  (.3)  of  a  foot  (on  a  large 
scale  drawing  it  was  found  to  be  .35)  from 
the  edge.  The  nearest  approach  to  the 
intrados  is  at  joints  3,  4  and  5  on  the  right, 
being  only  about  .7  to  .75  from  the  edges 
at  those  joints. 

Example  2  Draw  a  line  of  resistance  for  the  bridge, 
loaded  as  above,  through  the  lower  middle  third  limit  at  the 
left  springing,  the  upper  middle  third  limit  at  the  right 
springing  and  at  a  point  on  the  crown  joint  1. 1  ft.  above 
the  intrados.  It  passes  above  the  middle  third  limit  at  joint 
2  on  the  loaded  side,  its  maximum  departure,  and  just 
touches  the  lower  limit  at  joints  1  and  2  on  the  unloaded 
side. 

Example  3.  By  aid  of  formulas  (8),  (9)  and  (10),  draw  a 
line  of  resistance  through  the  lower  middle  third  limit  at 
the  left  springing  and  the  upper  middle  third  limit  at  joint 

2  under  the  load  and  also  through  the  upper  limit  at  the  the 
right  springing  joint. 

The  thrust  at  the  crown  will  be  found  now  to  act  0.76ft. 
below  the  centre  of  the  joint,  its  horizontal  component  being 
294  cu.  ft.  and  its  vertical  component  18  cu.  ft.  The  line 
of  resistance  everywhere  keeps  within  the  middle  third 
limits  except  at  joints  1  and  2  on  the  right  where  it  passes 
O.U  and  0.12  respectively  below  the  limit. 

Example  4.  A  load  of  13-3  tons  was  assumed  on  division, 

3  on  one  side,  and  it  was  found  that  a  curve  of  pressures 


49 

nmld  be  drawn,  for  this  .  (-centric  load,  within  the  inner 
third  of  the  arch  ring. 

If  the  backing  is  raised  higher,  thus  making  the  bridge 
weigh  more,  a  rolling  load  will  have  less  effect  upon  it . 
hence  a  less  depth  of  keystone  may  be  used.  Other  things 
{he  same,  it  is  a  simple  question  of  economy,  considering 
the  approaches,  whether  to  increase  the  height  of  surcharge 
above  the  arch  ring,  or  the  depth  of  the  arch  stones. 

Fig.  10  shows  the  effect  of  rolling  loads 
in  different  positions,  on  the  piers;  the 
middle  bay  not  being  loaded  but  with  its 
own  weight,  the  end  spans  as  per  figure. 
The  resultants  at  the  springing  joints  we 
have  before  determined;  combining  the 
two  on  any  pier  with  the  weight  of  pier, 
according  to  the  usual  rule  for  three  forces 
not  intersecting  in  one  point,  we  obtain  the 
final  resultants  on  the  bases  of  the  piers. 

It  is  seen  from  the  figure  that  the  40 
tons  on  both  sides  produces  a  more  hurtful 
effect  on  the  pier  than  a  40  ton  load  on 
one  side  only. 

By  combining  the  weight  of  abutment 
with  the  thrust  on  it,  we  find  that  the  cen- 
tre of  pressure  on  the  foundation  course  is 
sufficiently  within  the  limits  for  most  cases 
in  practice. 

The  dotted  line  in  the  abutment  gives 


50 


the  centres  of  pressure  of  all  the  forces 
acting  on  each  joint  for  the  joints  in  ques- 
tion. For  example,  to  find  where  this 
centre  of  pressure  is  on  the  springing 
line,  produced,  we  combine  the  inclined 
resultant  on  the  arch  joint  at  the  springing 
with  the  weight  of  the  abutment  above 
the  springing  line,  acting  through  its  cen- 
tre of  gravity.  This  resultant  makes  an 
angle  with  the  vertical  of  only  23°,  hence 
sliding  on  the  springing  course  is  not  to 
be  feared,  if  the  abutment  is  solidly  built. 

The  lines  of  resistance  as  drawn  for  the 
three  arches,  piers  and  abutments  are  not 
necessarily  the  true  ones. 

Further  on  will  be  given  a  method  of 
locating  approximately,  the  true  line  of 
resistance  for  a  well  built  arch,  with  thin 
mortar  joints,  between  immovable  abut- 
ments. The  abutments  in  the  figure  are 
of  such  proportions  as  to  be  practically 
immovable,  as  the  centre  of  pressure  on 
the  base  is  near  its  centre,  but  not  so  the 
piers.  The  first  pier  on  the  right  tends  to 
lean  to  the  left,  the  second  one  to  the 
right,  This  tendency  is  resisted  too  by  the 


central  arch,  which  thus  puts  forth  a 
stronger  horizontal  thrust  than  assumed, 
corresponding  to  a  line  of  resistance  pass- 
ing nearer  the  intrados  at  the  crown  and 
the  extrados  at  the  skewbacks,  the  maxi- 
mum efforts  being  produced  when  the  line 
of  resistance  passes  very  near  these  curves 
so  that  no  crushing  ensues.  We  are  not 
able  to  locate  this  line  with  our  present 
knowledge,  but  it  is  plain  that  this  central 
arch  will  put  forth  its  maximum  effort  if 
necessary,  to  prevent  much  motion  inwards 
of  the  tops  of  the  piers,  so  that  the  centres 
of  pressure  on  their  bases  will  not  depart 
as  far  from  their  centres  as  the  figure 
shows.  As  there  will  be  some  motion 
however,  it  tends  to  cause  the  line  of 
resistance  of  the  other  arches  to  travel 
down  the  skewbacks  at  the  piers  and  to 
move  up  the  crown  joints,  from  the  slight 
increase  of  span,  thus  giving  rise  to  a  less 
horizontal  thrust  from  those  arches,  which 
again  tends  to  correct  the  eccentricity  of 
the  thrust  on  the  piers.  If  preferred,  the 
lines  of  resistance  can  be  redrawn  in  these 
arches,  corresponding  to  a  minimum  thrust 


(within  reasonable  limits)  of  the  outer 
arches  and  a  maximum  thrust  of  the  cen- 
tral arch,  when  the  stability  of  the  piers 
will  be  more  apparent.  Experience  incli- 
,  cates  that  piers  of  the  proportions  shown 
are  perfectly  stable. 


CHAPTER  II. 

11.  A  MOKE  CORRECT  METHOD  OF 
MAKING  OUT  THE  TABLE  OF  WEIGHTS 
AND  CENTRES  OF  (TRAVITY. 

The  method  of  finding  the  weights  and 
centres  of  gravity  given  in  Art.  7,  although 
sufficiently  correct  for  flat  arches  with  a 
small  depth  of  key,  is  not  so  for  thick 
arches  approaching  the  semicircular  or 
elliptical  in  form.  The  following  is  sug- 
gested by  the  author  as  giving  all  desir- 
able accuracy  with  but  little  more  labor 
than  Scheffler's  method. 

The  arch  is  preferably  divided  into  a 
number  of  equal  voussoirs  (see  Fig.  11), 
and  the  vertical  lines  drawn  from  the 
upper  ends  of  the  joints  to  the  reduced 
contour  of  the  surcharge,  divides  the  latter 
into  trapezoids.  As  before,  we  draw  the 
medial  dotted  lines,  which  will  be  assumed 
to  pass  through  the  centres  of  gravity  of 
the  trapezoids,  though  the  latter  can  be 
found  exactly  by  the  usual  graphical  con- 


54 


struction  if  desired.  The  area  of  a  trape- 
zoid  —  width  X  mean  height  =:  w  v.  On 
multiplying  the  area  by  the  distance  from 
the  crown  to  the  medial  line  of  the  trape- 
zoid  (c)  we  have  the  moment  m  =  (w  v)  c« 
about  the  crown,  for  any  trapezoid. 

The  quantities  vr,  v,  s,  c  and  m  for  Fig. 
11  are  entered  in  the  table  below,  being  the 
upper  numbers  corresponding  to  the  joint 
given  in  the  first  vertical  column.  The 
corresponding  quantities  for  the  voussoirs 
are  the  lowest  numbers  of  the  horizontal 
rows. 

Calling  r,  in  Fig.  11,  the  radius  of  the 
extrados,  rl9  that  of  the  intrados  and  n  the 
proportion  of  the  circumference  included 
by  the  voussoir,  we  have  its  content 

7T(V'2 — T  2) 

—  for  a  thickness    of    1.     Now 
n 

this  is  equal  to     the  depth  (r — r^)  X  tne 

(1  T  — f-  V  \~1 

—  2  7t  -    — -)    hence  meas- 
n  'I     /J 

ure  the  middle  length  and  depth  on  a 
drawing,  their  product  will  give  the 
required  volume  of  a  voussoir  (=  2.35 
=  4.7  in  this  case). 


w 

v 

s 

c 

in 

S 

M 

C 

2.72 
2.35 

2.13 

2. 

5.79 
4.7 

1.38 
1.18 

7.99 
5.55 

10.49 

13.54 

1.29 

2.27 
2.35 

3.16 
2. 

7.17 

4.7 

3.88 
3.36 

27.82 
15.79 

22.36 

57.15 

2.55 

1.51 
2.35 

5. 

2. 

7.55 

4.7 

5.77 
5.03 

43.56 
23.64 

34.61 

124.35 

3.59 

.05 
2.35 

7.2 
2. 

3.6 
4.7 

6.77 
5.92 

24.37 

27.82 

42.91 

176.54 

4.11 

Tlie  centres  of  gravity  of  the  voussoirs 
will  be  assumed  to  lie  on  the  (dotted)  cen- 
tre line  of  the*  arch  ring  and  midway 
between  the  joints;  the  distances  from 
these  points  to  the  vertical  through  the 
crown  give  the  arms  in  column  (c).  The 
volume  (4.7)  of  a  voussoir  multiplied  by  its 
c,  gives  the  corresponding  ID  of  the  table. 

This  manner  of  considering  the  voussoirs 
and  surcharge  separately  is  continued,  until 
in  columns  S  and  M  the  quantities  referring 
to  the  same  joint  are  combined  by  the  con- 
tinued addition  of  the  quantities  in  col- 
ums  (s)  and  (ni)  respectively. 

If  the  voussoirs  are  taken  the  same  size, 
there  is  really  no  necessity  of  entering 
their  dimensions;  simply  giving  their  com- 
mon area  in  column  (s). 

When  the  voussoirs  are  taken  small 
enough  this  method  gives  all  desirable 
accuracy. 

Concentrated  loads  on  the  arch  are  easily 
included  by  introducing  a  third  row  of 
numbers,  for  any  voussoir  affected,  just 
above  those  given  for  the  trapezoids  as 
will  be  fully  explained  in  a  subsequent 
article. 


12.  A  CONVENIENT  METHOD  OF  DRAW- 
ING A  TKIAL  LINE  OF  RESISTANCE  IN  AN 

ARCH. 

Let  fig.  l'->  represent  a  semi-circular  arch 
of  Inn  ft.  span,  :>  ft.  key  and  o  ft.  depth  of 
surcharge  over  the  crown  of  the  same 
specific  gravity  as  the  voussoir.  The  live 
load  extends  from  the  crown  to  the  right 
abutment  and  weighs  .'),<•< M>  pounds  per 
foot  of  rails.  If  this  bears  on  a  width  of 
<]  feet  it  is  equivalent  to  a  layer  of  stone 
of  the  same  density  as  the  voussoir  (15<> 
Ibs.  pr.  cu.  ft.)  3.  4  ft.  high,  as  shown  in 
the  figure. 

The  spandrel  was  divided  up  by  verti- 
cal lines,  .">  ft.  apart  for  4"  ft.  from  the 
crown,  then  2  ft.  apart  for  the  next  10  ft., 
and  1  ft.  apart  for  the  remaining  3  feet. 
The  joints  1,  2,  3,  .  .  .  are  then  drawn 
as  in  the  figure. 

o 

The  following  is  a  condensed  table  of 
loads  (S  in  cubic  feet)  and  distances  from 
the  crown  to  their  centres  of  gravity  (C». 


58 


LEFT  HALF. 


Joints 

1 

2 

3 

4 

5 

6 

7 

S 

c 

30 

2.5 

62 
5.1 

100 

7.8 

146 

10.8 

201 
14. 

272 
17.4 

362 
21.1 

475 
25. 

Joints 

9 

10 

11 

12 

13 

819 
33.5 

14 

15 

16 

S 
C 

529 
26.6 

589 
28.3 

656 
30. 

732 
31.7 

869 
34.5 

925 
35.5 

1004 
36.8 

EIGHT  HALF. 

Joints 

1 

2 

3 

4 

5 

6 

7 

8 

S 
C 

47 
2.5 

96 
5. 

151 

7.7 

214 
10.5 

286 
13.6 

374 
16.8 

481 
20.2 

611 
23.9 

Joints 

9 

10 

11 

12 

13 

14 

15 

16 

S 

672 

739 

812 

895 

989 

1042 

1101 

1184 

C 

25.4 

27. 

28.6 

30.3 

32. 

33. 

33.9 

35.2 

60 

Let  us  pass  a  line  of  resistance  one- 
twelfth  the  depth  of  arch  ring  below 
the  centres  of  joints  8  and  the  crown 
joint,  or  through  a,  c  and  m.  By  the 
formula  method  of  art.  10  we  find  by  aid  of 
a  drawing,  &c.,  P  =  —  32. S,  Q  =  444.H. 
(It  will  be  instructive  for  the  reader  to 
test  these  values  by  the  purely  graphical 
method  of  art.  15,  which  is  generally  to 
be  preferred). 

From  m  lay  off  to  the  right  horizontally, 
i;  then  vertically  upwards, 
om  represents  the  resultant 
at  the  crown  joint. 

Lay  off  on  om  produced  mo  to  the  left 
and  equal  to  om.  Through  the  points  o 
thus  determined,  draw_verticals  and  lay 
off  from  o  the  distances  01,  02,  -  -  ,  equal 
to  the  values  of  S  pertaining  to  joints  1,2, 

,  as  taken  from  the  tables  pertaining 

to  the  right  and  left  sides  respectively. 
Straight  lines  from  m  to  1,  2,  -  -  ,  repre- 
sent the  resultants  of  the  thrust  at  crown 
and  load  down  to  joints  1,  2,  -  -  -  ,  in 
magnitude.  Their  positions  are  found  as 
follows :  draw  a  horizontal  through  ra,  and 


(51 


lay  off  on  it  the  numbers  in  column  C; 
the  first  table  referring  to  the  left  half  of 
the  arch,  the  last  table  to  the  right  half. 

From  the  points  so  found  draw  verti- 
cal lines  to  intersection  with  wo,  pro- 
duced if  necessary,  which  thus  give  the 
points  where  the  inclined  thrust  at  in  is 
to  be  combined  with  the  weight  from  the 
crown  to  any  joint,  to  find  the  resultant 
on  that  joint;  whose  intersection  with  it 
is  thus  the  centre  of  pressure  for  that 
joint. 

Thus  the  weight  from  the  crown  to 
joint  8  on  the  left,  acts  25'  to  left  of  m\ 
lay  off  25'  on  the  horizontal  through  m, 
then  drop  a  vertical  to  intersection  b  with 
mo;  then  draw  ~ba  \\  mS  of  force  diagram 
for  left  of  arch,  to  find  a  the  center  of 
pressure  for  joint  8.  Similarly  d  and  c 
are  found  for  joint  8  on  the  right.  These 
should  be  the  first  constructions  made  to 
test  the  values  of  P  and  Q  found,  which 
correspond  to  the  line  of  resistance  pass- 
ing through  a,  m  and  c. 

The  line  of  resistance  thus  drawn  pass- 
es below  the  middle  third  of  the  arch  ring 


62 


on  the  unloaded  side,  the  following 
amounts  in  feet:  at  joints  2,  3,  4,  .">  and  <), 
.o,  .4,  .3,  .2  and  .1  respectively;  it  then 
crosses  the  arch  ring,  passes  above  the  mid- 
dle third  about  joint  12,  and  cuts  the 
springing  joint  4.5  feet  outside  of  the  arch 
ring. 

On  the  loaded  side  it  passes  above  the 
middle  third  0.1  at  joints  4  and  5;  then 
crosses  the  centre  line  and  is  just  tangent 
to  the  lower  middle  third  limit  at  joint 
10,  below  which  it  again  crosses  the  arch 
ring  and  passes  into  the  abutment,  cutting- 
joint  10  about  3  feet  outside  of  the  arch 
ring. 

JZxercise.  Draw  a  line  of  resistance 
for  the  part  a  m  c  regarded  as  a  segrnen- 
tal  arch,  through  the  upper  middle  third 
limit  at  joint  8  on  the  left,  the  lower 
limit  at  joint  8  on  the  right  and  1,25  ft. 
above  the  intrados  at  the  crown. 

We  should  find  P=  -23.8,  Q  =  44'.U 
and  the  line  of  resistance  everywhere  keeps 
within  the  middle  third,  barely  touching 
the  lower  limit  at  joint  2  on  the  left  and 
passing  0.10  ft.  inside  of  the  upper  limit  at 


63 


joint  o  on  the  right  and  corresponding 
nearly  to  the  maximum  and  minimum  of 
tlie  thrust  in  the  limits  of  the  middle 
third  (see  Art.  20).  The  span  of  the  arch 
amc  is  75.45  feet,  the  rise  17.2  ft.  or 
between  %  and  V5  of  the  span. 

If  for  a  moment  we  regard  the  unaided 
semi-circular  arch  first  considered;  since 
the  line  of  resistance  (or  in  fact  any  line  of 
resistance  that  can  be  drawn  inside  the 
arch  ring  of  the  upper  portion)  passes 
outside  the  arch  ring  at  the  abutments,  the 
arch  will  fall,  the  parts  S-10  rotating  out- 
wards about  joints  K>  and  the  crown 
descending.  But  with  spandrels  built  of 
solid  masonry  up  to  about  joints  8  (called 
the  joints  of  rupture),  the  parts  8-16  can 
be  regarded  as  almost  immovable  and  the 
part  a  m  c  can  be  approximately  treated 
as  a  segimental  arch  on  fixed  abutments. 

However,  as  the  higher  the  abutments 
the  more  their  tops  will  yield  to  a  hori- 
zontal thrust,  the  depth  of  arch  ring 
determined  for  the  segimental  arch  a  m  c, 
regarded  as  resting  on  immovable  abut- 
ments, should  be  slightly  increased  to 


64 


allow  for  the  slight  horizontal  spreading 
at  a  and  c.  This  spreading  is  due  partly 
to  the  elastic  yielding  of  the  abutment 
from  8  down  to  the  foundation  and  partly 
to  the  closing  up  the  joints  of  the  rather 
fresh  mortar  in  the  vertical  joints  of  the 
spandrels  when  the  centres  are  struck. 

Some  constructors,  especially  the  French 
engineers,  increase  the  depth  of  arch  ring 
from  the  crown  to  the  abutments  so  that 
the  true  line  of  resistance  shall  not  leave 
the  middle  third  (or  other  limits)  any- 
where. This  is,  of  course,  the  best  way 
to  build  a  semi-circular  or  elliptical  arch, 
the  abutments  being  built  with  joints 
inclined  (about  at  right  angles  to  the 
thrust)  and  in  fact  treated  as  a  part  of  the 
arch  in  finding  the  true  line  of  resistance. 

13.  THE  USUAL  METHOD  OF  DRAWING 
A  LINE  OF  RESISTANCE.  EQUILIBRIUM 
POLYGON. 

In  Fig.  13,  representing  half  an  arch, 
suppose  the  thrust  S  at  the  crown  is  known 
in  position,  direction  and  magnitude  and 
that  the  weights  P1?  P2  and  P3  of  the  sue- 


cessive  voussoirs  01,  12,  '23  and  loads,  have 
been  found  and  laid  off  in  position  as 
shown.  From  some  convenient  point  O, 
draw  a  line  parallel  to  in  nQ9  the  direction 
of  the  thrust  S  at  the  crown  acting  at  m, 


and  to  the  scale  of  force  make  the  length 

00  equal  to  S. 

From  the  point  o  thus  found,  draw   a 
vertical  line  and  on  it  lay  off  successively 

01  =  ?!,  12  =  Pa  and  23  =  P8;  also  connect 
the  points   1,  2  and  3  with  O  by  straight 


lines.  These  lines  are  called  rays,  O  is  the 
pole  and  the  figure  is  known  as  the /*'>•/ w 
diagram.  The  rays  Ol,  O2,  O3,  represent 
the  resultants  on  the  joints  1,  ^',  '),  respect- 
ively in  magnitude  and  direction.  To 
find  their  position  on  the  arch,  produce 
the  thrust  S  at  m  to  intersection  n}  with 
P.  ;  from  this  point,  draw  a  line  parallel 
to  ray  Ol  of  the  force  diagram,  intersect- 
ing joint  1  at  <7;  produce  this  line  on  to 
intersection  <1  with  P2,  at  which  point 
draw  a  parallel  to  ray  O2,  intersecting 
joint  2  at  1> ;  again  produce  the  last  line  on 
to  intersection  f  with  P2,  at  which  point, 
draw  a  parallel  to  ray  O3  to  intersection  c 
with  joint  3.  The  points  </,  &,  c,  are  the 
centres  of  pressure  on  joints  1,  '2,  3  and 
a  broken  line  connecting  in  with  a,  a  with 
6  and  &  with  c  (dotted  line)  is  the  line  of 
resistance.  This  method  of  combining  each 
resultant  thrust  on  a  j  oint,  with  the  weight 
of  next  voussoir  and  load  to  find  the  result- 
ant on  the  next  lower  joint,  is  open  to  the 
objection  that  any  error  made  is  carried 
on,  whereas,  by  previous  methods,  any 
error  made  in  construction  is  confined  to 


Jie  joint  in  question.  The  polygon  m  ;/, 
If  is  called  an  equilibrium  polygon,  and 
we  note  that  it  does  not  coincide  with  the  line 
of  resistance;  also  that  it  passes  through 
the  centres  of  resistance  a  and  &,  but  does 
not  pass  through  c  ;  hence  an  equilibrium 
polygon  may  pass  out  of  the  arch  ring 
at  certain  points  and  yet  the  centres  of 
resistance  on  the  joints  be  found  in  the 
arch  ring;  so  that  it  cannot  be  used  alone 
in  testing  the  stability  of  an  arch  ring, 
particularly  for  semi-cicular  arches,  though 
it  is  generally  sufficiently  near  the  truth 
for  flat  arcs. 

If  we  produce  /'  c/,  which  gives  the 
direction  of  the  pressure  on  joint  2  to 
intersection  n2  with  in  ul  produced,  we 
have  n2,  a  point  in  the  vertical  through 
the  centre  of  gravity  of  the  part  of  the 
arch  from  the  crown  to  joint  '2.  This  is 
true,  because  we  know  that  we  must  com- 
bine S  at  the  crown,  with  the  weight  of 
arch  from  O  to  ^  (Pj+P2)  at  such  a  point 
on  the  line  of  action  of  S  as  to  give  a 
resultant  on  joint  ^  that  will  coincide  in 
position  and  direction  with  the  former 


C8 


resultant  d  /'.  The  only  point  that  satisfies 
this  condition  is  ng  which  proves  the  state- 
ment. Similarly/'  c  produced  to  inter- 
section with  m  n^  at  ?*3  gives  a  point  in 
the  vertical  passing  through  the  centre  of 
gravity  of  first  three  voussoirs  and  load. 

Conversely,  if  by  previous  methods  we 
compute  the  horizontal  distances  of  n,,  n2j 
n3,  from  the  crown  and  thus  fix  the  points 
n1?  n2,  ns  in  position,  and  then  find  the 
resultants  on  joints  1,  "2  and  3  by  drawing 
lines  from  n1?  ng  and  ns  parallel  to  rays  Ol, 
O2,  O3,  to  intersections  a,  b  and  c  with 
joints  1,  2  and  3,  these  resultants  produced 
to  intersection  will  form  sides  of  the 
equilibrium  ^polygon  m  nl  df. 

This  principle  will  be  utilized  in  apply- 
ing the  theory  of  the  solid  arch  further  on. 

(13a.)  Sometimes  the  position  of  the 
thrust  at  the  crown  is  not  given,  but  the 
thrust  at  the  lower  joint  3  (ray  O3)  is  given, 
passing  through  c.  In  this  case,  P,,  P2  and 
P3  having  been  laid  off  in  position  and  the 
force  diagram  constructed,  draw  from  c  a 
line  parallel  to  ray  O3,  to  intersection  f 
with  P3 ;  then  a  line  parallel  to  ray  O2, 


from  f  to  intersection  d  with  P2 ;  next  a 
line  from  d  parallel  to  ray  Ol,  to  intersec- 
tion na  with  P1 ;  lastly,  draw  from  ul  a  line 
parallel  to  ray  Go  =  S,  to  intersection  in 
with  crown  joint;  giving  thus  the  same 
equilibrium  polygon  m  7^1  df  as  before,  and 
the  same  line  of  resistance  m  a  b  c  deter- 
mined as  above. 

14.  SPECIAL  PROPERTIES  OF  THE  EQUILI- 
BRIUM POLYGON. 

Let  Fig.  14  represent  an  arch,  or  a  por- 
tion of  an  arch,  of  any  kind,  loaded  in  any 
manner,  the  joints  through  A  and  B  being 
any  two  joints  whatsoever,  and  the  joint 
through  I  being  any  intermediate  joint. 

Call  W  the  weight  of  arch  and  load  in- 
cluded between  joints  A  and  B,  W  and 
W"  the  weights  of  arch  and  load  included 
between  joints  A  and  I  and  joints  I  and  B 
respectively,  1'  =  horizontal  distance  from 
the  vertical  through  the  centre  of  gravity 
of  W  to  point  I.  Assume  the  thrust  upon  , 
the  joint  at  I  as  having  the  direction  Z  I 
H  and  at  the  point  H,  where  it  intersects 
the  vertical  through  the  centre  of  gravity 


70 

of  W,  draw  H  C  for  an  assumed  direction 
of  the  thrust  upon  the  joint  at  A,  the  point 


C  lying  upon  the  vertical  through  A  and 
either  above  or  below  it,  or  in  fact  coin- 
ciding with  it  as  a  special  case. 


71 

In  the  force  diagram  below  let  the  ver- 
tical Q  P,  to  any  scale,  represent  W,  Q 
G=  W",  and  G  P  =  W;  then  on  draw- 
ing a  line  through  P,  parallel  to  H  C,  and  a 
line  through  G  parallel  to  H  I,  their  inter- 
section gives  the  pole  O;  so  that  drawing 
through  Z  a  parallel  to  Q  O  to  intersection 
with  the  vertical  through  B,  we  establish 
the  point  D.  Connect  C  arid  D  by  a  straight 
line,  called  the  closing  line,  the  vertical 
through  I  meeting  it  at  E  and  the  perpen- 
dicular let  fall  from  I  upon  it  meeting  it  at 
X  Next,  through  the  pole  O,  draw  a  line 
O  M  parallel  to  C  I)  to  intersection  M  with 
Q  P  and  call  the  length  O  M  =  T  and  M 
P— V.  These  lengths  represent  the  two 
components  of  the  resultant  P  O  which 
acts  along  the  line  C  H  in  true  position. 
The  two  resultants,  whose  intensities  are  P 
O  and  O  Q,  acting  up  along  the  lines  C  H 
and  D  Z  respectively,  support  the  weight 
of  the  arch,  or  with  the  components  of  the 
weight  of  the  arch,  form  a  system  in  equil- 
ibrium. 

At  the  point  C  decompose  the  left  reac- 
tion into  two  components,  one  vertical 


acting  from  C  towards  S  at  a  horizontal 
distance  a  from  the  vertical  through  I,  and 
the  other  acting  from  C  towards  D. 

The  intensities  of  these  two  components 
are  given  by  the  lines  PM= V  and  OM  =  T 
in  the  force  diagram,  though  they  are 
represented  in  the  figure  above  to  a  dimin- 
ished scale. 

Now  if  we  suppose  the  part  of  the  arch 
to  the  right  of  the  joint  through  I  to  be 
removed  and  its  action  on  the  left  part  to 
be  replaced  by  the  resultant  pressure  it 
exerts  at  I  (  =  O  G  of  force  diagram)  act- 
ing to  the  le%  the  left  part  of  the  arch, 
extending  from  joint  I  to  A,  is  evidently  in 
equilibrium  under  the  action  of  this  force, 
the.  weight  W  of  this  part  AI  with  load 
and  the  two  components  V  and  T  at  C. 

Hence  taking  moments  about  I  of  th/jse 
forces  in  equilibrium,  we  have 

Va-WT  =  T.  IX. 

In  the  force  diagram,  the  line  ON  — H, 
the  pole  distance,  is  the  horizontal  distance 
from  O  to  PQ. 

Now  the  moment  T.  IX  is  equal  is  II. 


73 

IE;  since,  if  at  the  point  E  we  decompose 
T  (acting  along  CD)  into  vertical  and  hor- 
izontal components,  the  moment  of  the 
former  about  I  is  zero  and  the  moment  of 
the  latter =H.  IE,  which  is  thus  equal  to 
the  moment  of  T  about  I  =  T.  IX,  whence, 
Va-WT=H.  IE. 

Similarly  if  we  conceive  drawn  another 
equilibrium  polygon  A  K  J  Y  B,  with  pole 
O',  passing  through  the  points  A,  J  and  B 
in  the  verticals  through  C,  I  and  D  respect- 
ively, we  should  have 

Va-WT  =  H'  JF, 

where  V  is  the  component  AR  directed 
vertically,  T'  the  one  along  AB,  of  the 
left  reaction  acting  from  A  towards  K 
arid  H1  is  the  new  pole  distance. 

The  resultant  whose  line  of  action  is 
KJY,  is  the  resultant  on  the  joint  passing 
through  I  or  the  previous  joint  considered, 
J  being  simply  a  point  on  that  resultant 
in  the  vertical  through  I,  but  not  on  the' 
joint,  unless  the  latter  is  vertical.  It  fol- 
lows that  W  is  unaltered  and  it  remains 
to  be  proved  that  V=V. 


74 

We  have,  from  the  first  equilibrium  poly- 
gon the  system  of  forces  V  and  T  acting  at 
C,  the  weight  of  arch  and  load  W  acting  I  to 
left  of  vertical  B  D,  and  lastly  the  resultant 
reaction  at  D,  all  together  constituting  a 
system  of  forces  in  equilibrium.  Hence 
taking  moments  about  D,  we  have  (AL 
being  horizontal) 

V.  AL=W1. 

Similarly,  from  the  second  equilibrium 
polygon,  we  see  that  the  forces  V,  T',  W 
and  the  reaction  at  B,  constitute  a  system 
of  forces  in  equilibrium;  hence  taking 
moments  about  B,  we  have 

V'.AL=W1 

since  the  right  members  are  equal  in  the 
last  two  equations,  it  follows  that  V=V  ; 
whence  the  left  members  of  the  two 
equations  preceding  the  last  two  are  equal 
and  we  have 

IE 

H.  IE=H'.JF  .•.  H'  =  H  ^ 
j.b 

or  the  pole  distances  vary  inversely  as  the 
ordinates  IE  and  JF  from  I  and  J  to 
the  closing  lines  CD  and  AB. 


75 

The  above  conclusions  equally  hold  if 
the  weights  W,  W"  are  replaced  by  their 
components,  Representing  the  weights  of 
successive  voussoirs  and  loads,  and  the 
equilibrium  polygon  is  drawn  for  the 
entire  system,  since  the  pressures  on  joints 
A,  I  and  B  are  not  altered  by  the  decom- 
position ;  whence  the  previous  formulas  all 
hold  and  the  principles  in  question  are  es- 
tablished as  before. 

The  reader  who  is  acquainted  with  the 
principles  of  the  equilibrium  polygon  will 
recognize  that  C  H  Z  I)  is  the  equilibrium 
polygon  for  the  vertical  forces  V  (  =  PM 
at  C,  W  at  H,  W"  at  Z  and  a  vertical 
force  =  MQ  in  intensity,  at  B. 

Also  the  formula  above, 

Va-W'l'  =  H.  IE, 

proves  that  the  moment  of  all  the  vertical 
forces  to  the  left  of  a  point  I  is  exactly 
equal  to  the  pole  distance  H  measured  to 
the  scale  of  force,  multiplied  by  the  verti- 
cal distance  from  I  to  the  line  CD  (known 
as  the  closing  line)  measured  to  the  scale 
of  distance. 


76 

Similarly  since  V'=V,  the  same  moment 
is  equal  to  H'.JF  as  shown  above  or  the 
principle  is  generally  true  for  any  equili- 
brium polygon  drawn  as  above. 

As  in  drawing  a  line  of  resistance  it  is 
generally  most  convenient  to  know  the 
thrust  on  the  vertical  joint  U  at  the  crown 
in  position,  direction  and  magnitude,  the 
method  of  finding  it  for  the  polygon  whose 
pole  is  O',  will  now  be  indicated.  At 
the  intersection  of  JY  witli  the  vertical 
through  the  centre  of  gravity  of  the  por- 
tion of  the  arch  included  between  joints 
U  and  I  with  its  load,  combine  the  thrust 
at  J  (  =  G  O'  in  force  diagram),  acting  from 
left  to  right,  with  the  weight  of  the  por- 
tion considered =WG  on  force  diagram, 
giving  the  thrust  at  the  crown  =  O'W  ini 
magnitude  and  direction.  We  find  it  in 
position  by  drawing  through  the  above 
intersection  a  line  parallel  to  O'W  as 
as  shown  by  the  small  arrow.  From  this 
thrust  at  the  crown  we  draw  the  line  of 
resistance  right  and  left  of  the  crown  in 
the  usual  manner. 

If  the  joint  through    I  is   the    vertical 


77 

crown  joint,  then  KJY  is  at  once  the  line 
of  action  of  the  thrust  there,  its  magnitude 
being  equal  to  O'G  in  the  force  diagram. 

Where  the  joint  through  I  is  near  A,  it 
may  happen,  particularly  when  »the  tan- 
gent to  the  centre  line  of  the  arch  ring 
near  A  is  nearly  vertical,  that  the  vertical 
through  the  centre  of  gravity  of  W  or  the 
weight  from  joint  I  to  joint  A  with  load, 
lies  to  the  left  of  A. 

On  constructing  the  figure,  however,  it 
will  be  found  that  all  of  the  previous 
equations  hold,  so  that  the  above  con- 
clusions are  generally  true. 

The  principle  proved  above,  enables  us 
to  draw  an  equilibrium  polygon  through 
any  three  points,  as  A,  J  and  B  of  an  arch. 
To  do  this  take  I  (Fig.  14)  on  a  vertical 
through  J,  draw  the  joint  through  I  and 
find  the  values  and  positions  of  W  and 
W". 

Then,  as  detailed  above,  draw  the  trial 
equilibrium  polygon  C  H  I  Z  D  correspond- 
ing to  pole  O,  the  points  C  and  D  being 
in  the  verticles  through  A  and  B. 

On  drawing  from  O,  OM  ||  DC  to  inter- 


78 

section  M  with  QP  and  MO'  ||  AB,  a  dis- 
tance to  the  right  whose  horizontal  pro- 
jection is 

H-HIE 
JF, 

the  new  pole  O'  is  established. 

The  equilibrium  polygon  A  K  J  Y  B  can 
now  be  directly  drawn,  beginning  at  A,  J 
or  B  at  pleasure,  as  it  must  pass  through 
these  points.  Thus  beginning  at  J,  we 
draw  KY  ||  O'G  to  intersections  K  and  Y 
with  W  and  W";  then  lines  parallel  to 
PO'  and  O'Q  through  K  and  Y  respective- 
ly should  pass  through  A  and  B. 

15.  j&xample. 

In  Fig.  15  is  snown  an  arch  of  75  feet 
span,  15  feet  rise  and  7.5  feet  depth  of 
keystone,  the  top  of  the  backing  rising  to 
2  feet  above  the  top  of  arch  ring.  The 
specific  gravity  of  this  backing  is  supposed 
to  be  0.8  of  the  masonry  of  the  arch  ring, 
and  the  heights  above  the  arch  ring  are 
reduced  to  eight- tenths  of  the  original 
on  both  sides  of  the  arch,  though  it  is 


80 

only  shown  on  the  rignt  half  to  avoid 
confusion  of  lines.  Cooper's  class  "  extra 
heavy  A "  locomotive  is  placed  over  the 
left  half  in  the  position  shown.  The 
weight  on  each  pair  of  wheels  will  be 
supposed  to  bear  on  the  length  and  width 
of  a  cross  tie,  and  to  be  transmitted  verti- 
cally downwards.  If  we  regard  the  cross 
ties  as  8  feet  in  length,  the  weight  per 
foot  of  length  for  drivers,  is  15  -r-- 8  short 
tons  which  is  equivalent  in  weight  to  20.8 
cubic  feet  of  stone,  weighing  .07  ton  or 
140  Ibs.  per  cubic  foot.  This  supposes  the 
ring  stones  to  be  made  of  stone  of  this 
specific  gravity  which  corresponds  to  good 
sandstone  masonry. 

The  'equivalent  for  the  pilot  and  tender 
wheels  are  14.3  and  10.1  cubic  feet  respect- 
ively. These  weights,  resting  on  the 
successsive  voussoirs  through  the  span- 
drels, are  given  in  column  s,  Table  II 
(upper  numbers),  their  distances  from  the 
crown  are  given  in  column  c  and  their 
moments  about  the  crown  in  column  m. 
The  lower  numbers  pertaining  to  any 
t,  in  columns  s  and  m  refer  to  the  un- 


81 


loaded  half  and  are  found  from  Table  I  by 
summing  up  the  corresponding  quantities 
for  any  joint  which  are  made  out  as  ex- 
plained in  Art.  11. 

It  will  be  observed  that  the  division  of 
the  arch  ring  is  different  from  that  hitherto 
used.  As  we  shall  use  this  division  in 
a  subsequent  article,  from  considerations 
pertaining  to  the  theory  of  the  solid  arch, 
we  shall  state  that  the  centre  line  of  the 
arch  ring  is  divided  into  32  equal 
parts  (2.77  ft.  long  each),  and  then  the 
voussoir  joints  are  taken  in  succession  two 
divisions  apart,  except  for  the  two  vous- 
soirs  next  the  springing  and  the  two  next 
the  crown  where  only  one  division  is  taken. 
This  division  of  the  arch  ring  is  easily  made 
with  dividers.  The  volume  of  each  small 
voussoir  is  2.77x7.5  =  20.8  and  the  large 
ones  have  a  volume  =  2  X  20.8  ~  41.6  cubic 
feet.  The  horizontal  distances  from  the 
centre  of  each  voussoir  (taken  on  centre 
line)  to  the  crown  are  the  lowest  numbers 
of  column  c,  Table  1.  The  columns  S,  }.•: 
and  C  in  either  table  are  made  out  as 
usual  (see  Art.  11). 


TABLE  I. — RIGHT  HALF. 


Joints. 

\v 

v 

s 

c 

111 

8 

M 

C 

9 

2.93 

2.77 

1.6 

7.5 

4.7 

20.8 

1.46 
1.38 

29 

25.5 

36 

1.41 

10 

5.84 
5.54 

1.82 

7.5 

10.6 
41.6 

5.83 
5.48 

62 

228 

77.7 

326 

4.19 

11 

5.8 

2.48 

14.4 
41.6 

11.68 
10.97 

168 
456 

133.7 

950 

7.11 

12 

5.62 

3.6 

20.2 
41.6 

37.38 
36.32 

351 

679 

195.5 

1980 

10.13 

13 

5.5 

5.12 

28.2 
41.6 

22.93 
21.53 

647 
896 

265.3 

3523 

13.29 

14 

5.22 

7.07 

36.9 
41.6 

28.30 
26.57 

1044 

llo.> 

343.8 

5672 

16.49 

15 

4.9 

9.4 

46.1 
41.6 

33.36 
31.33 

1538 
1303 

431.5 

8513 

19.75 

16 

4.fi3 

12.1 

56.0 
41.6 

38.10 
35.88 

2134 
1493 

529.1 

12140 

22.95 

17 

2.22 

14.4 

32.0 
20.8 

41.57 
38.03 

1330 
7(.»1 

581.9 

14261 

24.  30 

581.9 

14261 

83 


TABLE  II. — LEFT  HALF. 


s 

0 

c 

m 

S 

M 

C 

8      25.5 

36 

25.5 

36 

1.41 

14.3 

7    52.2 

4.15 

59 
290 

92.0 

385 

4.19 

2(5.8 
6     5(i. 

12.25 

328 
624 

174.8 

1337 

7.65 

26.8 

o     61.8 

18. 

482 
11)30 

263.4 

2849 

10.82 

26.8 

4: 

22.5 

603 
1543 

360.0 

4995 

13.87 

26.8 
3 

27. 

724 
2149 

465.3 

7868 

16.90 

16.1 

•J     87.7 

34.1 

549 
2841 

569  .  1 

11258 

19.78 

16.1 
1     H7.6 

38.9 

626 
3627 

682.8 

15511 

22.72 

0     52.8 

2121 

735.  C, 

17632 

23.97 

735.6 

17632 

All  the  computations  in  these  and  sim- 
ilar subsequent  tables  were  made  with  a 
in. inch  slide  rule,  which  ensures  suf- 
ficient accuracy  with  but  small  mental 
wear  and  tear. 


84 


If  it  is  desired  to  pass  a  line  of  resistance 
through  the  points  A  and  B  at  the  spring- 
ing joints  and  through  J  on  the  5th  joint, 
we  first  draw  the  load  line  C  870  ...  o  for 
the  left  half  of  arch  and  lay  off  on  it  from 
Table  II,  column  S,  08=25,  07  =  'X>, 
etc.;  also  from  column  C  lay  off  the 
successive  distances,  on  a  horizontal 
(dotted)  line  through  the  crown,  to  the 
verticals  through  the  centres  of  gravity  of 
the  loads  from  the  crown  to  any  joint. 
Similarly  C'  '.>,  .  .  .  ,  17  is  the  load  line  for 
the  right  half  and  Y  Z  is  a  vertical  through 
its  centre  of  gravity. 

We  now  assume  the  thrust  at  the  crown 
to  act  along  same  line  HZ  and  draw  rays 
CO  and  C'O'  parallel  to  it;  at  the  point  H 
where  this  line  meets  the  vertical  through 
the  centre  of  gravity  of  the  left  half  of 
the  arch  and  load,  draw  H  A.  Then  draw 
a  ray  from  o  in  left  load  line,  parallel  to 
II  A  to  intersection  with  ray  C  O  at  O, 
thus  fixing  the  pole  O.  Make  C'  O'  =  C  O 
to  fix  the  right  pole,  since  C  O  gives  the 
magnitude  of  the  thrust  at  the  crown. 

The    thrust   at  crown    meets  Y  Z  at  Z5 


85 


from  which  point  draw  a  parallel  to  ray 
O'  —  IT  to  intersection  D  with  the  vertical 
through  B.  Connect  A,  D  and  A,  B  and 
mark  the  points  E  and  F  where  they  are 
intersected  by  a  vertical  through  J. 

At  the  point  where  the  trial  crown  thrust 
meets  the  vertical  through  the  centre  of 
gravity  of  arch  and  load  from  crown  to 
joint  5,  10.^  ft.  to  left  of  crown  (Table 
II)  draw  a  parallel  to  ray  Oo  to  intersection 
I  with  the  vertical  through  J. 

Through  the  trial  pole  O  (as  in  Art.  14) 
draw  O  M  ||  A  D  to  intersection  M  with 
load  line ;  then  draw  M  P  ||  A  B  a  distance 
to  the  right  whose  horizontal  projection  is, 

T  K 

H'  =  H  As  the  distances  I  E  and  J 

J  r . 

F  are  rather  short,  double  them  and  lay 
of  along  same  line  C  R  through  C,  C  L  = 
'2  x  JF,  CH  =  '2  x  IE.  Then  if  Sis  the 
intersection  of  a  horizontal  line  through  C 
and  a  vertical  through  O,  connect  L  and  S 
and  draw  R  Q  parallel  to  L  S  to  intersec- 
tion Q  with  C  S  produced,  which  construc- 
tion gives  H'  =  C  Q.  A  vertical  through 
Q  to  intersect  M  P  at  P  gives  the  new  pole 


86 

P.  At  the  right,  draw  ray  C'  P '  parallel 
and  equal  to  a  ray  from  P  to  C  (not  drawn) 
and  we  have  the  new  pole  P'  at  the  right. 

We  have  only  now  to  draw  through 
J  a  line  parallel  to  ray  Po  to  intersection 
T  with  the  line  of  action  of  the  weight 
from  the  crown  to  joint  5,  to  fix  a  point  T 
in  the  line  of  action  of  the  new  thrust  at 
the  crown. 

Through  T  draw  a  line  parallel  to  PC, 
and  from  the  intersections  K  and  Y  of  this 
line  with  the  verticals  through  H  and  Z 
draw  lines  parallel  to  rays  Go  and  O'-IT 
respectively,  which  lines  should  pass 
through  A  and  B  if  the  work  has  been 
done  correctly. 

If  this  does  not  obtain  the  error  is  per- 
haps largely  due  to  not  taking  off  the 
lengths  IE  and  JF  (with  dividers,  never 
with  scale)  with  sufficient  accuracy.  At 
any  rate,  the  construction  must  be  repeated 
if  necessary  until  the  line  of  resistance  will 
pass  through  the  three  points  A,  B  and  J. 

The  centre  of  resistance  on  any  joint  is 
found  in  a  similar  manner  to  that  already 
given  for  the  springing  joint  B.  The 


87 


broken  line  connecting  the  centres  of 
resistance  on  all  the  joints  is  shown  by 
the  clotted  line  to  pass  near  the  centre  line 
throughout,  which  is  due  to  the  great 
depth  of  arch  ring  chosen  in  this  case  for 
the  sake  of  a  clear  figure. 

When  the  point  J  is  on  the  crown 
joint,  take  I  to  coincide  with  it  and  draw 
the  line  of  action  HZ  of  the  trial  thrust  at 
the  crown  through  I  =  J.  The  construc- 
tion then  proceeds  as  before,  only  the 
final  thrust  at  the  crown,  K  Y,  is  now 
drawn  through  1= J  parallel  to  ray  PC. 

The  construction  is  thereby  simplified 
for  this  case. 

The  graphical  methods  given  above  of 
determining  completely  the  thrust  at  the 
crown  for  a  line  of  pressures  passing 
through  any  three  points  in  the  arch  ring 
will  probably  be  preferred  to  the  analytical 
methods  of  Art.  10. 

The  preceding  article  indicates  the  entire  construction 
for  the  case  represented  by  Fig.  15.  Let  the  reader  repeat 
this  construction  on  a  scale  of  3  or  4  feet  to  the  inch  and 
compare  with  the  numerical  values  of  the  components  of 
the  thrust  at  the  crown  found  by  the  formula  method. 

As  another  example,  in  the  arch  shown  by  Fig.  9  (Art. 
10)  use  the  quantities  in  the  Tables  of  Arts.  8  and  9  and 


88 


pass  a  line  of  pressures  through  the  lower  middle  third 
limit  at  the  left  springing  and  through  the  upper  middle 
third  limits  at  joint  2  on  the  left  and  at  the  springing  joint 
on  the  right.  With  a  scale  of  30  feet  to  the  inch  it  is 
found  that  the  thrust  at  the  crown  acts  0.76  foot  below 
the  centre  of  the  joint,  its  horizontal  component  being 
;  294  cu.  ft.,  vertical  component  18  cu.  ft. 

At  joints  1  and  2  on  the  right  the  centres  of  resistance 
fell  0.14  and  0.12  respectively  below  the  middle  third  limit. 


89 


CHAPTER  III. 

CURVES  OF  RESISTANCE  CORRESPONDING 
TO  THE  MAXIMUM  AND  MINIMUM  HOR- 
IZONTAL THRUST.  UNIT  STRESSES. 

16.  PROPERTIES  OF  CURVES  OF  RESIST- 
ANCE. 

In  Fig.  16,  e,  c  a  represents  an  unsymmet- 
rical  arch,  or  an  arch  acted  on  by  forces 
not  symmetrical,  vertical  or  inclined. 

Let   P— resultant  of  the  external  forces 


acting  on  the  arch  between  a  and  c,  not 
including  the  reaction  R  at  a.  Then  on 
combining  R=a&  with  P,  we  get  the 
centre  of  pressure  c  on  the  joint  cc^ 
Similarly  we  could  proceed  for  other 
points,  b9  d,  e,  of  the  curve  of  resistance, 
corresponding  to  the  resultant  R=.o&7 
acting  through  a  in  the  direction  ak. 


90 


Let  al  b  cl  d  e^  be  a  second  curve,  cor- 
responding to  the  reaction  R'  at  a^  Now 
if  S  is  such  a  force,  acting  towards  the 
left,  that  when  combined  with  R,  it  gives 
R'  as  a  resultant,  we  can  find  a  point  cl9 
on  joint  cc,,  of  the  new  curve  of  resistance, 
either  by  combining  R'  with  P  as  before, 
or  by  combining  its  components  with  P : 
thus  call  the  resultant  of  R  and  P,  T;  this 
combined  at  I  with  S,  gives  a  resultant 
which  cuts  joint  c  c^  at  t\,  a  point  lying 
between  Id  and  c,  Id  being  in  the  direction 
of  s  produced. 

By  this  construction  it  is  seen  that  the 
new  curve  of  resistance,  corresponding  to 
the  reaction  R'  at  a',  passes  through  b  and 
d,  the  points  where  k  I  intersects  the  first 
curve  of  resistance;  for  other  joints,  as  eeiy 
the  new  curve  lies  nearer  kl  than  the  first 
curve;  since  when  S  acts  to  the  left,  the 
combination  of  T,  for  any  joint,  with  S, 
gives  a  resultant  acting  between  T  and  S, 
which  therefore  cuts  the  joint  nearer  kl 
than  the  first  center  of  pressure. 

The  above  supposes  that  neither  R  nor 
S  are  vertical,  but  that  both  act  to  the 


91 


left,  whence  the  horizontal  component  of 
R'  exceeds  that  of  R.  The  joints  are, 
moreover,  not  supposed  inclined  more  than 
!K)°  from  the  vertical,  counting  from  the 
top. 

17.  Prop.  If  two  curves  of  resistance 
cut  each  other,  the  curve  which  lies  nearest 
the  straight  line,  which  joins  their  common 
points,  corresponds  to  the  greatest  hori- 
zontal thrust. 

We  have  seen  in  the  preceding  article 
that  the  two  curves  can  only  intersect  on 
the  straight  line  k  I  (Fig.  10)  as  implied  in 
the  proposition. 

Now  if,  at  any  joint  c  c},  the  centre  of 
pressure  cl9  corresponding  to  the  curve 
al  b  cl  del9  lies  nearer  kl,  the  straight  line 
joining  b  and  d,  than  the  curve  abcde, 
then  we  may  suppose  a  force  S,  acting  in 
the  direction  kl,  to  be  combined  with  T 
at  I,  to  effect  it.  The  force  S,  thus  found, 
must  therefore,  when  combined  with  R  at 
a  give  R';  since  R  and  S  produce,  the 
same  effect  as  R';  so  that  all  points  of  the 
first  curve  can  be  found  by  combining  R 
with  the  resultant  of  the  force  P,  up  to 


the  joint,  and  afterwards  combining  their 
resultant  with  S. 

The  force  S,  acting  to  the  left,  increases 
the  horizontal  component  of  the  resultants 
on  each  joint;  hence  the  curve  a^bc^de^ 
corresponds  to  a  greater  horizontal  thrust 
than  the  curve  abed e,  as  stated  in  the 
proposition. 

If  the  arch  is  symmetrical,  the  curves  of 
pressure  are  symmetrical  with  respect  to 
the  crown,  whence  kl  must  be  horizontal. 

18.  (1).  If  a  cure e  of  resistance  has  two 
points  common  to  the  intrados  and  an  in- 
termediate point  common  to  the  extrados, 
it  corresponds  to  the  minimum  horizontal 
thrust. 

For,  suppose  the  curve  a  b  c  d  e^  Fig.  16, 
touches  the  extrados  near  c,  the  intrados 
on  both  sides  nearer  the  abutments. 

Then  any  other  curve  of  resistance, 
al  b  cl  d  e^  that  remains  in  the  arch  ring, 
must  cut  the  first,  only  in  points  on  the 
straight  line  kl,  joining  any  two  points 
of  intersection. 

Now  the  new  curve,  near  the  points  of 
contact  of  the  first  curve  with  the  con- 


tour  curves  of  the  arch  ring,  must,  if  it 
remains  in  the  arch  ring,  pass  nearer  kl 
than  the  first  curve,  whence,  by  Prop. 
Art.  17,  the  first  curve  corresponds  to  a 
less  horizontal  thrust.  Q.E.D. 

(2).  If  (f  <<f<rre  of  resistance,  a  b  c  dc, 
Fig.  17,  has  two  points  of'  contact,  b  and 
'7,  with  the  extrados,  and  <m  intermediate 
point  of  contact  c  -with  the  intrados,  it 
corresponds  to  a  iidninnwn  horizontal 
thrust,  if  b  c  dy  in  the  vicinity  of  c,  lies 


between    the    intrados    and    the     straight 
line  bd, 

For  any  other  curve,  lying  in  the  arch 
ring,  as  the  dotted  curve,  must  lie  nearer 
the  straight  line  kl,  joining  their  points 
of  intersection,  than  the  first,  in  the  vicinity 
of  b  c  and  d,  and  thus  corresponds  to  a 
greater  horizontal  thrust.  This  case  of 


94 

the    minimum  is    rarely    or   never    found 
in  practice. 

(3).  If)  however ,  the  intrados,  in  the 
vicinity  of  c  lies  between  the  curve  b  c  d, 
Fig.  18,  and  the  straight  line  bd>  the 
curve  corresponds  to  <i  imurinnuu  hori- 
zontal thrust ;  since  this  curve  lies  nearer 
Id  than  any  other,  as  the  dotted  curve  of 
resistance. 


It  is  seen  that  Id  in  Fig.  18  lies  above 
bd,  whereas  the  reverse  occurs  in  Fig.  17. 

When  a  curve  of  resistance  possesses 
both  the  properties  of  the  maximum  and 
minimum  of  the  thrust,  the  arch  is  at  the 
limit  of  stability;  as  see  all  the  figures 
relating  to  the  experiments  in  the  Appen- 
dix. The  above  principles  were  first  stated 
by  Dr.  Herman  Schetfier. 

I'.*.  When  the  arch  is  symmetrical  (and 
symmetrically  loaded),  the  curves  of  resist- 


95 

ance  are  symmetrical  with  respect  to  the 
vertical  through  the  crown;  hence  for  a 
half  arch  and  load  we  can  state  the  follow- 
ing propositions : 

(1).  When  the  point  of  contact  with 
the  extrados  is  higher  than  the  point  of 
contact  with  the  intrados,  the  curve  of 
resistance  corresponds  to  the  minimum 
horizontal  thrust. 

(:i).  When  the  point  of  contact  with 
the  extrados  is  lower  than  the  point  of  con- 
tact with  the  intr ados,  the  curve  of  resist- 
ance corresponds  to  the  maximum  horizon- 
tal thrust.  It  generally  touches  the  intra- 
dos  at  the  crown  joint. 

The  curve  corresponding  to  the  minimum 
thrust  generally  touches  the  extrados  at 
the  crown  for  usual  loads  and  depths  of 
arch  ring;  but  for  thin  arch  rings  with 
little  or  no  surcharge  above  the  crown, 
especially  with  gothic  arches,  the  curve  of 
resistance  passes  below  the  crown  and 
touches  the  extrados  at  some  lower  point. 
This  likewise  may  happen,  even  for  seg- 
mental  arches  of  usual  depth,  when  a  heavy 
load  is  placed  over  the  middle  of  either 


96 


haunch,  as  shown  in  Art.  9  for  the  curves 
drawn  corresponding  to  the  minimum  hori- 
zontal thrust  within  the  middle  third 
limits. 

For  segmental  arches,  the  lower  point 
of  contact  for  either  maximum  or  minimum 
thrust  is  generally  at  the  springing  joint. 

:><).  The  conclusions  above  hold  equally 
when  we  wish  to  find  the  maximum  or 


minimum  thrust  for  curves  contained  with- 
in the  middle  third  or  any  other  limits, 
only  we  substitute  the  upper  and  lower 
limiting  curves  for  extrados  and  intrados 
in  the  enunciations. 

In  Fig.  19  the  dotted  line  represents  a 
curve  of  resistance  corresponding  to  the 
maximum  and  minimum  of  the  thrust  at 
the  same  time,  within  the  limits  shown. 

The  part  a  b  c  corresponds  to  the  max.  and 


97 

bed  to  the  min.  of  the  thrust  within  those 
limits;  for  b  lies  above  a  straight  line 
drawn  from  a  to  c,  Art.  18  (3),  and  c  lies 
between  b  and  d,  Art.  18  (1). 

In  an  arch  by  itself,  if  but  one  curve  of 
resistance  can  be  drawn  within  its  contour 
curves,  thus  corresponding  at  once  to  the 
maximum  and  minimum  thrust,  the  arch 
is  at  the  limit  of  stability. 

UNIT  STRESSES  AT  ANY  POINT  OF  A  JOINT. 

21.  In  Figure  3  of  Article  5,  the  result- 
ants of  the  molecular  stresses  on  the  joints 
ab,  at  b1?  a2  b2,  are  Q,  R15  R2,  respectively. 
To  find  these  stresses  at  any  point  of  a 
joint  as  a2  b2.  The  resultant  R2  on  the 
joint  a.,  b2  meets  it  at  A2,  whose  distance 
from  the  nearest  edge  ( a2  A2  in  this  case)  we 
shall  call  d.  The  joint  is  rectangular  in 
shape ;  its  width  perpendicular  to  the  plane 
of  the  paper  being  unity,  and  its  radial 
length  a2  b2  we  shall  call  h. 

As  R2  is  generally  inclined  to  the  nor- 
mal to  the  joint,  resolve  it  at  A2  into  two 
components,  the  first,  which  call  R,  being 
normal  to  the  joint  and  the  other  acting 


98 

parallel  with  the  joint.  The  last  compo- 
nent is  a  shearing  force  and  undoubtedly 
lessens  the  resistance  of  the  joint,  but 
because  its  influence  is  difficult  to  estimate 
it  is  generally  neglected. 

It  will  first  be  assumed  that  the  mortar 
in  the  joint  can  withstand  both  tension  and 
compression,  in  which  case  R2  can  fall  out- 
side the  joint  a2  b2  (Fig.  :>),  a  distance  d 


^r«*   B 

r/£.  kt 

""FE^— 

>• 

^(    C7 

' 

/  i    P* 

+  Pv    A        -  R 

I  : 

f 

1 

e  

iH     r—  ^C^ 

! 

,  

D 

d 

R_  

.  i 

measured  along  the  line  of  the  joint  pro- 
duced, and  still  stability  be  assured  if  the 
mortar  possesses  sufficient  strength.  Simi- 
larly for  concrete  arches.  In  Fig.  *?<>,  let 
B  D  represent  the  joint  a2  b2  of  Fig.  o  and 
at  the  point  where  the  resultant  on  this 
joint  cuts  the  joint  or  the  joint  produced, 
resolve  it  into  shearing  and  normal  compo- 


99 

nents ;  neglecting  the  former,  we  have  the 
normal  component  R,  acting  through  a,  a 
distance  d  -f-  J£  h  from  the  centre  E  of  the 
joint  (Fig.  20). 

At  the  point  E  conceive  two  opposed 
forces -{-R,— R,  equal  and  parallel  to  R  to 
act.  This  does  not  destroy  equilibrium. 
To  avoid  confusion  +  R  and— R  are  drawn 
through  A,  a  point  in  the  normal  to  BD  at 
E,  but  it  is  understood  that  A  is  supposed 
to  coincide  with  E.  The  force  R  with  the 
force— R  forms  a  right-handed  couple  RR, 
that  can  be  replaced  by  the  equal  couple 
pp  or  the  forces,  £,£'....  c',  c,  equal  and 
opposed  to  the  uniformly  increasing  ten- 
sible  resistances  from  E  to  B  and  the 
compressive  resistances  from  E  to  D,  E 
lying  in  the  centre  of  gravity  of  the  cross- 
section. 

From  the  theory  of  flexure,  calling  t  =  c 
the  stress  per  square  unit  at  the  extreme 
edge  B  or  D,  we  have  the  moment  of  the 
couple  RR  =  M  =  R  (d  +  %  h)  =  V6  ch2 ; 
whence, 


100 

The  remaining  force  +  R  at  E  (A),  acting 
at  the  centre  of  gravity  of  the  joint  corres- 
ponds to  a  uniform  compression, 

h 

over  the  whole  joint  (shown  by  the  little 
arrows  to  the  right,  though  really  acting 
along  B  D). 

As  the  forces  given  by  (1)  and  (2)  act 
at  the  same  time,  their  algebraic  sum  or 
"  effect "  (see  figure)  gives  the  actual 
stresses  along  joint  B  D,  which  are  thus 
seen  to  be  uniformly  increasing  from  a 
neutral  axis,  not  passing  through  E.  The 
tension  at  B  =  t— rand  compression  at  D 

As  in  voussoir  arches  the  resultant  rarely 
or  never  passes  outside  the  arch  ring,  let 
us  suppose  hereafter  that  it  cuts  the  joint 
to  which  it  refers,  in  its  interior  a  distance 
d  from  the  nearest  edge ;  then  we  have, 

_6R(^h-d) 

h2 
R 


101 
whence, 

,-,=,(,--)«.,..«. 
.+,=,(,-«)•....<» 

As  this  theory  supposes  that  the  limit  of 
elasticity  has  been  nowhere  exceeded,  the 
stretch  or  shortening  of  the  "fibres"  is 
proportional  to  the  stress  and  therefore  to 
the  distance  from  the  neutral  axis;  hence 


a  plane  section  before  strain  remains  a 
plane  section  after  strain  as  in  the  ordinary 
theory  of  beams. 

It  is  seen  from  (3)  that  t  —  r  is  positive, 
or  tension  is  exerted  at  B,  for  d  <  K  h ; 
for  d  ==  K  h,  t  —  r  =  o  or  there  is  no 

stress  at  B  and  the  stress  at  D  =  c  -(-  r  =  2 

T> 

r-  or   double   the   average  (see  Fig.   21)* 


102 

lastly,  for  d  >  K  h,  the  stress  on  the  joint  is 
compressive  throughout  (Fig.  22). 

Hence  when  the  resultant  cuts  the  joint 
within  its  middle  third  there  are  only  com- 
pressive forces  exerted  on  the  joint;  when 
it  passes  outside  the  middle  third,  tensile 


Fig.  22 


forces  are  brought  into  play  if  the  mortar 
can  supply  them. 

In  the  last  case,  if  the  mortar  cannot 
resist  tensile  forces,  the  normal  component 
of  the  resultant  will  be  decomposed  into 
stresses,  over  a  length  of  joint  3d,  pro- 


103 

portional  to  the  ordinates  of  a  triangle  and 
the  part  of  the  joint  beyond  this  length 
(od)  will  open. 

This  is  plain,  because  we  have  just  found 
that  for  d  =  }3  h,  that  od— h,  was  under 
compression,  the  stress  at  the  farthest  edge' 
from  the  resultant  being  nothing. 

The  stress  at  the  most  compressed  edge 
is  likewise  double  the  average  on  the  part 
of  the  joint  under  compression  .*.  it  is 

:3  - '-  for  all  cases  where  the  resultant  lies 
-}  d 

in  an  outer  third  of  the  arch  ring  and 
the  mortar  cannot  resist  tension. 

The  last  formula  and  formula  (4)  suffice 
to  give  the  maximum  compression  per 
square  unit  for  the  cases  to  which  they 
refer,  where  the  voussoirs  are  each  in  one 
block.  In  case  the  arch  ring  is  made  of 
several  rolls  as  in  brickwork  or  when  we 
meet  with  joints  transverse  to  the  radial 
joints  as  we  proceed  along  the  latter,  the 
above  theory  is  only  approximately  true. 

Example  1.  In  an  arch  of  5  feet  radial  length  of  joint 
at  the  springing,  the  resultant  has  a  normal  component  of 
73.36  tons  and  it  acts  1  ft.  below  the  centre  of  the  joint. 


104: 

(1).  What  is  the  stress  at  the  intrados  if  the  mortar 
cannot  resist  tension  ? 

Answer,  "xS   ',''-=  32.6  tons  pr.  sq.  ft. 

(2).  What  is  the  stress  at  the  intrados  and  extrados 
if  the  mortar  can  resist  tension? 

Here  B=73.3G.  h=5ft...  d  =1.5  ft.;  hence  by  eqs.  (:J) 
and  (4), 

Stress  at  extrados=2/ri~^iH)!5i=  2.0. 
V  5/5 

Stress  at  intrados  =2(2  —    '        J ^-^—)  =  32. 2 
tons  per  square  foot. 

Example  2.  In  the  example  above,  if  the  resultant  acts 
0.5  ft.  above  the  centre  of  the  joint,  find  the  unit  stivsst-s 
at  the  intrados  and  extrados. 


X'2.  Iii  the  theory  of  the  solid  arch, 
which  will  be  presently  referred  to,  it  is 
necessary  to  know  the  moment  M  of  the 
resultant  on  any  joint  with  respect  to  the 
centre  of  that  joint.  This  may  be  ex- 
pressed in  three  different  ways:  first,  by 
the  product  of  the  resultant  by  the  perpen- 
dicular from  the  centre  of  the  joint  upon 
it;  second,  by  R  ( }.,  h — d  )  as  above,  since 
the  shearing  component  has  no  moment, 
and  lastly  by  multiplying  the  horizontal 
component  of  the  resultant  on  any  joint  by 
the  vertical  distance  from  the  centre  of 
the  joint  to  where  a  vertical  through  this 


105 

centre  meets  the  resultant.  This  is  plain, 
since  at  the  intersection  of  the  vertical 
with  the  resultant  decompose  the  latter 
into  vertical  and  horizontal  components. 
The  moment  of  the  former  about  the 
centre  of  the  joint  is  zero;  hence  the1 
moment  of  the  latter  (the  constant  hori- 
zontal thrust  of  the  arch )  is  equal  to  that 
of  the  resultant  itself. 


•i-).  METHOD  OF  FAILURE  OF  ARCHES. 

In  the  appendix  will  be  found  an  ac- 
count of  a  number  of  experiments  on  small 
wooden  arches  at  the  limit  of  stability 
with  their  corresponding  resistance  lines, 
which,  of  course,  correspond  to  the  maxi- 
mum and  minimum  thrust  at  the  same 
time,  within  certain  limits.  (See  art.  '2  '.) 

In  the  fourth  experiment,  irith  a  yield- 
UKJ  pier,  the  top  of  the  pier  and  the 
haunches  of  the  arch  moved  outwards  and 
the  crown  descended.  In  this  case  the 
limiting  line  of  resistance  (for  the  slightly 
deformed  arch)  touches  the  extrados  at  the 
crown,  the  intrados  at  the  haunches  and 


106 

cat  the  outside  edges  of  the  piers.  This  is  the 
method  of  failure  of  any  symmetrical  arch 
witli  yielding  abutments. 

With  rigid  abutments  the  method  of 
failure  for  gothic  or  sequential  arches,  with 
a  load  at  the  crown  is  given  by  experi- 
ments one  and  eleven,  and  for  an  eccentric- 
load  by  experiments  sixteen  and  seventeen. 
The  figures  show  the  lines  of  resistance-. 
The  .arches  rotated  about  those  edges 
•\vhich  the  lines  of  resistance  approached 
nearest.  As  the  arches  were  slightly  de- 
formed at  the  instant  before  rupture,  if  the 
resistance  lines  had  been  drawn  for  the  de- 
formed arch,  they  would  necessarily  have 
passed  through  the  edges  of  the  joints,  as 
no  crushing  there  was  experienced  for 
these  very  light  wooden  arches. 

When  arches  having  rigid  abutments 
are  built  with  too  thin  an  arch  ring,  it  is 
found  that  the  arch  fails  by  the  crown 
rising  and  the  haunches  falling  inward. 
TJie  line  of  the  centres  of  pressure  passes 
through  the  intrados  at  the  crown  and 

o 

abutment  joints  and  touches  the  extrados 
;it  the  haunches,  the  segments  of  the  arch 


107 

rotating  about  these  edges;  the  upper 
segment  turns  upwards  about  the  extrados 
edge  at  the  haunches  and  the  lower 
segment  falls  inwardly,  rotating  about  the 
springs.  The  spandrels,  if  any,  must 
crack  over  the  springs,  haunches  and 
crown;  the  lower  segments  fall  inwardly 
into  the  void  between  the  abutments  and 
are  speedily  followed  by  the  upper  part 
with  the  spandrels. 

There  are  doubtless  some  arches  in  exist- 

|  ence    now,    perilously   near   the    kind    of 

I  rupture  just  outlined.  Woodbury,  in  his 
treatise  on  the  ar,ch,  has  devoted  consider- 

'  able  attention  to  this  latter  mode  of  failure. 
He  supposes  the  arch  unloaded,  the  span- 

sdrel  filling  to  be  of  the  same  density  as  the 
voussoirs  and  to  be  carried  up  level  with 

I  the  crown  and  the  arch  ring  to  be  uniform 

mi  thickness.  He  found  that  a  semi-circu- 
lar arch  whose  depth  was  1/53  of  the  span, 

'could  just  contain  one  line  of  resistance 
and  no  more  within  the  limits  of  the  arch 
ring  down  to  the  lower  joints  of  rupture, 
and  that  when  the  span  was  2(J  times  the 

(depth  of  key,  only  one  line    of   resistance 


108 


could  be  inscribed  within  the  middle  third 
drawn  to  the  lower  joints  of  rupture.  The 
lower  joints  of  rupture  made  angles  with 
the  vertical  of  07°  in  the  first  case  and 
about  fJ'-i0  in  the  second. 

If  we  call  s  =  span,  h  =  rise,  and  k=. 
depth  key,  the  following  table  gives 
Woodbury's  limits  of  k  in  terms  of  the 
span  for  segmented  arches  of  various 
ratios  of  Tito*;  the  first  set  of  values 
k  -=-  s  giving  the  ratios  of  depth  of  key  to 
span  which  permits  only  one  line  of  resist- 
ance to  be  drawn  within  the  limits  of  the 
arch  ring;  the  second  set  of  values,  the 


h 

1 

1 

1 

1 

1 

1 

s 

4 

5 

6 

7 

8 

10 

1 

k 

1 

1 

1 

1 

1 



8 

47 

50 

M 

00 

65 

78 

k 

1 

1 

1 

1 

1 

1 

s 

25 

•Jf> 

•21 

30 

33 

39 

ratios  of  k  to  s  in  order  that  only  one  line 
of  resistance  can  be  inscribed  within  the 
middle  third  of  the  arch  ring.  If  the 
spandrel  is  carried  above  the  crown,  these 


109 

ratios  will  become  less;  but,  if  after  the 
centres  are  struck,  the  spandrels  are 
brought  to  a  level  with  the  top  of  key- 
stone, the  last  ratios  should  certainly  never 
become  less  or  joints  will  open.  In  fact,  if 
these  values  are  attained  the  construction 
for  the  solid  arch  will  give  a  line  of  resist- 
ance passing  slightly  outside  of  the  middle 
third  and  thus  bringing  tensile  stresses  on 
fresh  mortar  at  some  of  the  joints.  Proper- 
ly, the  spandrels  should  be  built  up  pro- 
gressively from  key  to  abutment,  so  that 
the  height  at  the  key  is  attained  before 
that  at  the  abutment.  As  it  will  be  well 
for  the  reader  to  test  some  of  these  values, 
it  may  be  mentioned  that  for  the  first  set 
of  values  above,  the  line  of  resistance 
touched  the  intrados  at  the  crown,  the  ex- 
trados  at  the  haunches,  and  the  intrados  at 
the  springing.  The  curves  limited  to  the 
middle  third  touched  the  lower  limit  at 
the  crown  and  springing  and  the  upper 
limit  about  the  haunches. 

Exercise. — When  the  span  is  100  feet, 
rise  20  feet,  and  depth  of  key  4  feet 
( V25  space),  and  the  spandrel  rises  to  a 


110 

level  from  the  top  of  key,  construct  the 
single  line  of  resistance. 

The  method  of  failure  of  segmental 
arches  with  rigid  abutments  and  an  eccen- 
tric load  over  the  haunch  of  the  left  half 
may  be  illustrated  by  a  reference  to  fig. 
19.  Conceive  the  arch  ring  to  diminish  in 
depth  so  that  finally  but  one  curve  of  re- 
sistance can  be  drawn  therein.  It  will  be 
found  to  touch  the  extrados  to  the  left  of 
the  crown  and  at  the  right  springing 
joint;  it  will  touch  the  intrados  at  the  left 
springing  joint  and  a  little  to  the  right 
of  the  crown. 

In  a  large  arch,  crushing  at  the  edges  is 
experienced  before  this  minimum  depth  of 
arch  ring  is  attained.  In  any  case  the 
arch  will  sink  at  the  joints  under  the  load, 
the  joints  at  the  intrados  opening,  whereas 
the  arch  ring  rises  a  little  to  the  right  of 
the  crown,  since  the  pressure  there  is  near- 
ly all  concentrated  at  the  lower  edge.  At 
the  left  springing,  the  lower  part  of  the 
arch  rotates  downwards  about  the  lower 
edge  and  at  the  right  springing  about  the 
upper  edge  of  the  joint.  As  a  consequence, 


Ill 

the  arch  divides  into  three  parts;  the  left 
part  falling  inwards,  the  middle  portion 
rising  at  the  right  but  falling  at  the  left 
end  and  the  right  segment  rotating  up- 
wards about  the  upper  edge  of  the  right 
springing  joint. 

In  any  kind  of  an  arch,  loaded  in  any 
manner,  the  method  of  failure  is  easily  ar- 
rived at  by  simply  studying  the  line  of 
resistance  pertaining  to  the  case. 


112 


CHAPTER  IV. 

LENE  OF  RESISTANCE  DETERMINED  AS  IN 
A  SOLID  ARCH.  METHOD  OF  ISOLATED 
LOADS  FOR  SEGMENTAL  ARCHES.  COM- 
PUTATION OF  DEPTH  OF  KEYSTONE. 

'^4.  A  great  many  approximate  solutions 
have  been  proposed  for  the  voussoir  arch, 
but  none  satisfactory.  The  true  line  of  re- 
sistance in  an  arch  depends  primarily  upon 
its  elasticity,  and  likewise  upon  the  care 
with  which  the  stones  are  cut  and  fitted, 
the  thickness  and  yielding  of  the  mortar 
joints,  the  settlement  and  time  of  striking 
of  the  centres  if  the  mortar  joints  have  not 
hardened,  and  finally  the  yielding  of  the 
piers  or  abutments.  So  many  of  these  in- 
fluences cannot  be  exactly  estimated  that 
the  author  has  hesitated  about  applying 
the  theory  of  the  solid  arch  "  fixed  at  the 
ends"  to  the  voussoir  arch,  particularly  on 
account  of  its  complexity,  though  111  Van 
Nostrand's  Magazine  for  January  and  No- 
vember, 1871),  he  claimed  that  the  theory 


113 

of  the  solid  arch  was  the  most  exact  solu- 
tion for  the  cases  assumed,  and  a  graphical 
treatment  was  given  in  the  last  named  ar- 
ticle, to  which  reference  will  be  made  fur- 
ther on. 

From  Prof.  Swain's  article  in  Van  Nos- 
trand's  Magazine  for  October,  1880,  it  is 
to  be  inferred  that  the  application  of  the 
theory  of  elasticity  to  the  stone  arch  had 
already  been  considered  by  a  few  authors 
mentioned.  In  187!)  Winkler  published 
his  notable  theorem  (given  in  the  article 
last  mentioned) ;  also  Castigliano  applied 
the  theory  of  the  solid  arch,  after  the 
method  of  "  least  work,"  to  stone  arches.  In 
the  same  year  Prof.  Greene,  of  the  Uni- 
versity of  Michigan,  published  a  more 
practical  treatment,  founded  on  the  ana- 
lytical theory  of  the  circular  arch,  using 
the  method  of  isolated  loads. 

In  this  chapter  a  method  similar  to  Prof. 
Greene's  is  used,  the  tables,  however,  be- 
ing obtained  by  aid  of  Winkler's  tables,  for 
the  solid  arch  "  fixed  at  the  ends."  The 
xp<r/i  of  the  centre  line  of  the  arch  ring 
was  divided  into  equal  parts,  and  the  quan- 


114 

titles  in  the  tables  found  for  isolated  loads 
at  the  points  of  division  by  interpolation 
(by  aid  of  diagrams  to  a  large  scale),  from 
Winkler's  constants,  which  refer  to  equal 
angular  divisions,  though  a  few  direct  com- 
putations were  made  for  loads  very  near 
the  abutment.  Winkler's  theory  and  tables 
for  the  solid  arch  are  given  in  Du  Bois's 
"  Graphical  Statics,"  and  the  preliminary 
computations  of  cx  and  c2  were  made  out  as 
explained  in  "  Theory  of  feolid  and  Braced 
Arches"  by  the  writer,  p.  00,  the  deforma- 
tion of  the  arch  ring,  due  to  bending  mo- 
ments, being  alone  considered.  The  gen- 
eral table  given  in  this  chapter  refers  only 
to  arches  whose  rise  is  one  fifth  of  the  span. 
The  method  of  equal  horizontal  divisions 
adopted  here  offers  great  practical  advan- 
tages, and  enables  one  with  small  labor, 
comparatively,  to  investigate  the  strength 
and  stability  of  a  given  stone  arch.  The 
position  of  the  lice  load  causing  maximum 
departures  of  the  centres  of  pressure  from 
the  centres  of  the  joints,  is  more  accurately 
ascertained  than  hitherto,  and  some  unex- 
pected conclusions  were  established. 


115 

25.  The  theory  of  the  solid  arch  "  fixed 
at  the  ends,"  is  strictly  applicable  to  a 
solid  arch  of  stone,  iron  or  other  material 
perfectly  fitted,  when  not  under  stress,  to 
rigid  abutments,  the  theory  requiring  three 
conditions  to  be  satisfied,  viz.:  (1)  the  tan- 
gents to  the  centre  line  at  the  springs  are 
fixed  in  position,  (2)  span  invariable,  and 
(o)  the  vertical  displacement  of  one  spring- 
ing above  the  other  equals  zero;  all  having 
reference  to  the  deformation  of  the  arch 
ring  due  to  the  stresses  caused  by  its  own 
weight  and  any  load  that  may  be  ap- 
plied. 

The  theory  is  evidently  rigidly  applica- 
ble to  a  voussoir  arch,  with  no  mortar  in 
the  joints,  provided  the  voussoirs  are  cut  so 
perfectly  that  the  arch  fits  accurately  be- 
tween the  rigid  abutments  when  not  under 
stress.  When  thin  cement  mortar  is  used 
in  the  joints  and  allowed  to  harden  before 
the  centres  are  struck,  the  conditions  are 
but  little  altered ;  but  for  bad  fitting  stones 
and  thick  mortar  joints,  not  sufficiently 
hardened,  neither  this  theory  nor  any  other 


116 

is  exactly  applicable,  though  it  may  be  re- 
garded as  some  guide  in  the  relative  di- 
mensioning of  arches  of  various  spans. 

It  must  be  carefully  observed,  too,  that, 
if  in  a  voussoir  arch,  without  mortar,  fitting 
perfectly  when  not  under  stress  between 
the  skewbacks,  the  centre  of  pressure  on 
any  joint,  as  determined  by  the  theory  of 
the  solid  arch,  falls  without  the  middle 
third,  the  joint  will  only  bear  on  a  length 
equal  to  three  times  the  distance  from  the 
centre  of  pressure  to  the  nearest  edge  of 
the  joint  (Art.  21).  In  this  case  only  the 
bearing  surface  of  the  joint  must  be  inclu- 
ded in  the  formulas  for  fixing  the  resist- 
ance line.  Hence  a  new  determination  has 
to  be  made  on  this  basis  and  so  on,  until 
the  assumed  and  computed  bearing  joints 
agree.  This  method,  although  possible,  is 
very  tedious,  so  that  the  theory  is  only  a 
practicable  one  when  the  arch  ring  is 
such  that  the  true  line  of  the  centres  of 
pressure  lies  everywhere  within  the  middle 
third. 

Only  such  arches  are  examined  in  this 
work,  besides  space  does  not  permit  the 


117 

consideration  of  any  arch  rings  except 
those  of  uniform  cross-section. 

20.  In  Fig.  *239  the  circular  curve,  of 
rise  equal  to  one-fifth  of  its  chord  or  span, 
represents  the  centre  line  of  an  arch  ring 
of  constant  section.  The  span  is  5  units 
in  length,  the  rise  one  unit.  Each  half 
span  is  divided  into  10  equal  parts  and 
vertical  lines  drawn  through  the  points  of 
division.  Where  the  successive  lines  cut 
the  curve  will  be  designated  as  points  <>, 
1,  '2,  :3,  .  .  .  ,  10  and  1',  •>',  .r,  .  .  .  10' 
for  the  left  and  right  halves  respectively, 
the  left  springing  point  being  called  lo, 
the  right  springing  10',  the  crown  0  and 
the  numbers  increasing  regularly  along 
the  arc  from  the  crown  to  the  two  spring- 
joints  respectively. 

If  a  single  load  W  is  placed  on  the  arch, 
supposed  to  be  without  weight,  its  equili- 
brium polygon  will  consist  of  two  straight 
lines.  When  W  is  directly  over  a  point 
of  the  arch  previously  fixed,  as  4,  the 
table  gives  at  once  c^  y  and  ca  in  terms  of 
h  =  rise  of  centre  line  of  arch  considered, 
where  cl  =  vertical  distance  from  point  10 


118 


119 

on  arc  to  side  of  equilibrium  polygon,  c2 
the  same  for  point  10'  on  arc  and  y  —  ver- 
tical distance  above  the  crown  to  apex  D 
of  equilibrium  polygon. 

Thus  if  W  acts  at  4  on  arch,  the  rise  of 
whose  centre  line  (above  the  centre  of  the 
springing  joints)  is  10,  the  span  50,  we 
have  from  table,  that  the  resultant  at  left 
abutment  acts  —  .10  X  10  or  l.G  below 
centre  of  joint,  the  resultant  at  right 
springing,  acts  -f-. 290  X  10  or  2.90  above 
centre  of  joint,  and  the  apex  D  lies  .21.0 
X  10  =  2. 10  above  the  highest  point  of 
centre  line  of  arch  ring,  plus  coefficients 
corresponding  to  points  above  the  arc  and 
minus  below,  as  just  indicated.  On  laying 
off  W  on  a  vertical  line  just  to  left  of  arch 
as  shown,  and  drawing  lines  from  the  ex- 
tremities parallel  to  the  sides  of  the  equi- 
librium polygon  passing  through  D,  the 
intersection  gives  the  pole  of  the  force 
diagram.  The  length  of  the  horizontal 
line  from  the  pole  to  the  line  representing 
W,  gives  H  =  horizontal  thrust  due  to  W 
alone,  and  the  line  divides  W  into  the  two 
vertical  components  of  the  reactions  Vt 


120 


tf 


PQ 


W 

II 

a" 


w 


+-f      M    1    !    II 


-  x  x  Ci  c;  H 


Ct  71  — *-  l^  CC  b-  X  t^  tr:  i 
•'  — <  ^C  O  t»  ^  r:  Tl  C  : 
COiCOOa>^HQOOCOb"' 


7i  cc  »-f  ^i  r:  o  i~  O  O  : 

•4  I  f  I  I  —  • 


121 

and  V2  at  left  and  right  springings  re- 
spectively. For  W  =  1  (to  a  large  scale) 
we  can  thus  find  graphically  the  co- 
efficients in  columns  H,  V,  and  Y2,  but  it 
is  better  to  compute  them  from  easily 
derived  formulas: 

V,::^H     ;.V,=  ^H; 

X  '>  -  X 

H  =  wWAD  +  Bm 
\     x        5—  x/ 

We  have  here,  A  D  =  1  -{-  y  —  (cj, 


paying  attention  to  the  sign  of  c,.  Since 
x  is  known,  H  can  be  at  once  computed 
and  afterwards  Vj  and  V2.  The  coefficients 
thus  found  are  to  be  multiplied  by  W  in 
any  application,  as  indicated  in  the  table. 
The  moment  about  any  point  of  the 
centre  line  of  the  arch  ring,  for  any  weight 
W  is  equal  to  the  H  corresponding,  multi- 
plied by  the  vertical  distance  from  the 
point  to  the  equilibrium  polygon  corres- 
ponding to  W  (see  Art.  22).  The  alge- 
braic sum  of  such  moments  due  to  any 
number  of  weights,  gives  the  total  moment 


122 

at  the  point  and  the  sum  of  the  IPs  gives 
the  total  horizontal  thrust  clue  to  the 
weights.  If  the  moment  at  the  point,  due 
to  the  weight  of  the  arch  is  found  and 
added  to  the  preceding  moment  and  its  H 
likewise  added  to  the  previous  sum  of  the 
IPs,  then  the  quotient  of  the  total  mo- 
ment divided  by  the  total  II,  gives  the 
vertical  distance  from  the  point  on  the 
centre  line  of  the  arch  ring  to  the  equili- 
brium polygon  corresponding  to  the  weight 
of  arch  and  loading  considered.  This  is 
the  method  to  be  used  in  fixing  any 
point  of  the  equilibrium  polygon  after 
the  theory  of  the  solid  arch  "  fixed  at  the 
ends".  When  the  moment  is  plus,  the 
equilibrium  polygon  is  above  the  centre 
line  of  the  arch  ring  at  the  point;  when 
minus,  below. 

For  convenience,  the  values  of  M,  = 
Hc1  and  M2  =  Hc2  ( the  moments  at  the 
left  and  right  springing  joints)  as  well  as 
Mj-j-Mg  are  given  in  the  general  table. 
The  coefficient  for  a  weight  at  the  crown 
(  point  0  of  arc)  is  only  half  of  Mt  -f-  M2 
for  reasons  that  will  appear  directly. 


123 


Let  fig.  24  represent  any  arch  whose 
rise  is  one-fifth  the  span;  the  rise  of  ite 
centre  line  is  also  one-fifth  its  chord  from 
centre  to  centre  of  springing  joints.  Di- 
vide the  half  chord  of  the  centre  line  into 
ten  equal  parts  and  erect  verticals  at  the 
points  of  division.  Reduce  the  length  of 


124 

the  part  of  these  verticals  comprised  be- 
tween the  extrados  of  the  arch  and  the 
horizontal  roadway  line  to  eight  tenths  of 
each.  This  is  best  done  graphically. 
Thus  if  e  g  is  laid  off  equal  to  10  and  ef 
equal  to  8  to  any  scale,  and  we  lay  off  any 
length  from  e  along  e  g  and  from  its  ex 
tremity  draw  a  parallel  to  g  f  to  intersec- 
tion with  6/',  the  distance  from  this  inter- 
section to  e  gives  the  reduced  length.  The 
material  above  the  arch  ring  to  the  reduced 
contour  can  then  be  regarded  as  weighing 
the  same  per  cubic  foot  (.07  ton)  as  the 
arch  ring.  On  drawing  dotted  verticles 
half  way  between  the  first  verticals,  the 
area  of  the  trapezoids  comprised  between 
any  two  successive  dotted  verticals  will  be 
equal  to  the  width  multiplied  by  the  length 
of  the  full  vertical  between  them,  and  it 
will  be  regarded  as  a  force  acting  along 
the  medial  (or  full)  vertical.  In  fact,  for 
a  length  of  arch  equal  to  unity,  this  area 
is  also  the  volume  of  a  prism  having  for  a 
base  this  area,  and  by  multiplying  by  .07  it 
can  be  reduced  to  tons. 

In  all  the  computations  below  the   por- 


125 

tion  of  the  arch  from  either  springing  joint 
to  the  first  dotted  vertical  will  be  neglected, 
as  its  influence  is  very  small  in  fixing  the 
true  equilibrium  curve.  The  weight  at  the 
crown  is  that  due  to  the  portion  comprised 
between  the  adjacent  dotted  verticals  on 
either  side.  This  division  of  the  arch  is 
different  from  that  hitherto  used. 

27.  Let  us  proceed  now  to  the  consider- 
ation of  an  arch  of  100  ft.  span.  The  rise 
is  20  ft.  and  the  depth  of  the  key  5  ft.,  the 
horizontal  roadway  rising  2  ft.  over  the 
crown.  The  first  column  in  the  table  be- 
low gives  the  joint  of  the  arch  at  which 
the  weight  is  concentrated.  The  "  depth" 
of  a  "  trapezoid  "  (  column  2)  multiplied 
by  the  constant  "  width  "  5.2  (  column  3  ) 
gives  the  area  =  volume  ==  W,  expressed 
in  cubic  feet.  We  have  only  to  multiply 
these  values,  as  well  as  those  given  in  col- 
umns H  and  M,  by  .07  to  reduce  to  tons 
when  desired. 

The  coefficients  of  columns  H  and  M, 
-f-  M2  are  taken  from  the  general  table 
above.  On  multiplying  these  by  the  suc- 
cessive values  of  W  we  get  the  horizontal 


126 

thrusts  and  moments  given  in  columns  H 
and  Mt  of  the  table.  Any  thrust  in  col- 
umn H  is  that  due  to  the  load  at  the  point 
corresponding,  hence  the  sum  gives  the 
total  thrust  due  to  loads  0,  1,  2,  -  -  -  ,  !». 
On  adding  to  this  the  same  sum,  less  that 
due  to  the  load  at  the  crown,  we  get  the 
thrust  due  to  the  weight  of  the  entire  arch 
=  51*4.0  cubic  feet,  since  the  loads  at  equal 
distances  from  the  crown  are  the  same. 
Similarly  M,  for  load  at  0  -f-  Mt  for  load 
at  O1,  is  the  same  as  Mj  for  load  at 
<>  +  M2  for  load  at  6  =  (  M4  +  M2) 


Table  for  100  ft.  span,  20  ft.  rise,  5  ft.  depth  of  Key. 


"S 
o 

"ft 

<& 

g 

3 

W 

H 
Coeff 

H 

M^Mo 
Coeffs." 

MI 

PL| 

ft 

£ 

cu.  ft. 

cu.  ft. 

0 

6.6 

5.2 

34.3 

1.1639 

39.92 

-f-.  1769 

+  6.07 

1 

6.8 

35.4 

1.1512 

40.75 

+  .3489 

+  12.35 

2 

7.2 

37.4 

1.0864 

40.63 

+  .2944 

+  11  01 

3 

8. 

41.6 

.9807 

40.80 

+  .2089 

+  8.69 

4 

9.1 

47.3 

.8473 

40.08 

+  .1152 

+  5.45 

5 

10.5 

54.6 

.6867 

37.49 

+  .0075 

+  0.41 

6 

12.2 

63.4 

.5098 

32.32 

—  .0973 

—  6.17 

7 

14.4 

74.9 

.3327 

24.92 

—  .1780 

—13.33 

8 

17.2 

89.4 

.1706 

15.25 

—  .2143 

—19.16 

9 

20.3 

105.6 

.0497 

5.25 

—  .1688 

—17.83 

583.9                     317.41                         —12.51 

549.6                     277.49                                     20 

Weight  of  arch  =  11  33.  5            H=594.9                M1=—  250.2 

127 

for  load  at  6.  The  same  principle  holds 
for  any  two  loads  at  equal  distances  from 
the  crown.  The  coefficient  for  the  load  at 
the  crown  was  not  doubled,  as  there  is  no 
other  load  corresponding  to  it. 

On  adding  up  the  figures  in  the  last 
column  and  multiplying  by  h  =  2()1  we 
find  the  total  moment  —  —  250.2,  and  on 
dividing  this  by  the  total  H=  504.0  we 
find  that  the  resultant  at  the  left  springing 
passes  0.42  foot  below  the  centre  of  the 
joint.  The  same  holds  at  the  right  spring- 
ing on  account  of  symmetry.  The  equili- 
brium polygon  can  now  be  drawn,  as  the 
vertical  component  = :  J  weight  of  arch 
—  500.75  and  H  =  51)4.0  are  given  as  well  as 
Cj=  — .42.  Construct  the  force  diagram  by 
drawing  Hand  from  its  left  extremity ;  layoff 
successively  on  a  vertical  downwards  half 
the  load  at  crown  and  the  loads  at  1,  2, 
-  -  -  ,  0  (see  fig.  24).  We  draw  the  equil- 
ibrium polygon  from  a  point  .42  below  the 
centre  of  the  left  springing  joint,  as  ex- 
plained in  Art.  (13a).  It  is  very  near  the 
centre  line  and  is  shown  approximately  in 
fig.  24.  Its  vertical  distance  from  points 


128 

l>,  o,  0  and  o'  on  the  arc  are  respectively 
+.27,  +.2, — .25  and  —  1 ;  so  that  multiply- 
ing by  H  =  5(J5,  the  moment  at  these 
points  are  -f-  101,  -\-  120,  —  141)  and  —  00 
respectively.  We  have  previously  seen 
(Art.  13)  that  the  resultant  along  in  n,  say 
of  fig.  24,  is  strictly  that  pertaining  to  the 
joint  between  m  and  n  at  a,  the  true  re- 
sistance curce  passing  slightly  below  rn\ 
still  for  purposes  of  comparison  below  it 
is  near  enough  to  consider  the  line  m  n 
to  represent  the  line  of  action  of  the  result- 
ant acting  on  the  joint  passing  through 
the  point  where  the  vertical  through  m 
cuts  the  centre  line,  particularly  as  we  shall 
find  that  the  maximum  departure  of  the 
line  of  resistance  from  the  centre  of  the 
joints,  when  the  live  load  is  considered,  is 
nearly  always  at  a  springing  joint  where 
no  error  is  made.  Further,  as  the  result- 
ants on  the  upper  joints  are  nearly  perpen- 
dicular to  them  for  usual  loads  the  inter- 
section of  a  perpendicular  from  m  on  the 
joint  corresponding  can  be  regarded  as 
the  centre  of  pressure,  for  purposes  of 
comparison  below. 


We  are  now  prepared  to  consider  the  ad- 
ditional influence  of  the  live  load,  which  in 
all  the  subsequent  examples  in  this  chapter 
will  be  taken  as  a  locomotive  load  of  Oooo 
pounds  per  foot  of  track,  20  feet  in  length 
or  slightly  greater  or  less,  corresponding  to 
the  horizontal  divisions  of  the  arch, 
followed  by  a  tender  load  of  2400  Ibs.  per 
ft.  30  ft.  long,  about,  and  this  followed  by 
another  locomotive  load  as  before.  This 
about  corresponds  to  Cooper's  class  extra 
heavy  A,  without  the  pilot  wheel.  For 
cross  ties  S  ft.  long,  these  loads  for  a  slice 
of  the  arch  I  foot  thick,  are  750  and  30o 
Ibs.  per  ft.  respectively.  As  the  horizontal 
divisions  of  the  arch  are  5.2  ft.  each,  the 
locomotive  load  on  each  division  =  750  x 
5.2  Ibs.,  or  the  weight  of  27.8  cubic  ft.  of 
stone  weighing  140  Ibs.  per  cubic  ft.;  the 
tender  load  on  each  division  is  11.1  cubic 
ft.  stone.  The  first  locomotive  will  be  as- 
sumed to  cover  4  divisions  of  5.2  ft.  each 
or  2o.8  ft. ;  the  tender  6  divisions  or  31.2  ft. 
As  M  and  H  both  vary,  for  any  point  as 
we  shift  the  live  load,  it  is  only  by  trial 


130 

that    the    maximum    value  of  M^-H=c, 
can  be  found  for  the  point  considered. 

Thus  consider  point  5  of  centre  line  of 
arch  ring  where,  for  dead  load  only  we 
have  found  M  =  -f-120  and  H  =  5(,>4.(,». 
On  a  large  scale  drawing  similar  to  fig.  23 
(except  that  loads  at  all  points  1,  2,  3  -  -  - 
are  considered),  on  measuring  the  vertical 
ordinates  from  point  5  of  arc  to  the  equili- 
brium polygons  corresponding  to  weights 
W=27.8  at  4,  5,  0,  7,  and  adding  the  results 
(  =  .778)  we  have  the  total  moment  due 
to  locomotive  load  at  4,  5,  6,  7,  =  .778  X  h 
W  =  .778  X  20  X  27.8  =:  -f  432.0.  That 
due  to  tender  load  at  8,  9,  is  similarly  = 
.061  X  20  X  11.1  =  +  13.5 ;  adding  moment 
=  -\- 120  due  to  dead  load,  the  total 
moment  at  point  5  =  -f-  500.1.  The  hori- 
zontal thrust  due  to  loads  27. 8  each  at  4,  5? 
0,  7,  is  found  by  adding  the  coefficients  in 
column  H  of  general  table  for  points  4,  5, 
0,  7,  and  multiplying  by  27.8  .-.  2.37G5  X 
27.8  =  00.1.  Similarly  for  tender  load  at 
points  8,  t),  the  thrust  is  .2203x11.1  =  2.4. 
Adding  the  thrust  due  to  weight  of  arch, 
-VJ4.1)  to  the  sum  of  these  two  and  we  have 


131 

the  total  horizontal  thrust  =  663.4.  On 
dividing-}- 5 6 6.1  by  this,  we  find  that 
the  equilibrium  polygon  due  to  dead  load, 
locomotive  load  at  4,  5,  (>,  7,  and  tender 
load  at  8,  9,  passes  0.89  foot  above  the 
centre  of  the  joint  through  point  5  on  arc, 
whence  on  drawing  a  normal  to  the  joint 
through  the  point  found,  the  resist- 
ant is  found  to  pass  0.8  above 
centra  measured  along  the  joint.  The 
live  load  may  now  be  moved  to  right 
or  left  one  or  more  divisions,  but  for  no 
other  position  is  the  resultant  on  joint  5  so 
far  from  the  centre  as  is  found  by  a  com- 
putation similar  to  the  above.  A  similar 
investigation  for  point  6  with  locomotive 
loads  at  4,  5,  0,  7  and  tender  loads  at  8,  9, 
gives  M  =  +520,  H  =  W>4,  so  that  c  =  + 
.7S  or  less  than  for  point  5.  From  fig.  23 
it  is  evident  that  for  points  5  or  6  the  live 
load  should  not  extend  as  far  as  point  2 
from  the  left,  as  the  moment  due  to  a  load 
at  2  or  to  the  right  is  negative.  Trial 
shows  that  for  a  maximum  c  at  6  the  load 
should  not  extend  to  o,  but  from  10  to  4  in- 
clusive as  just  given. 


132 

At  the  crown  the  max.  departure  is 
caused  by  loads,  say  from  4  to  10  and  41  to 
lo1 ;  but  it  is  not  so  great  as  elsewhere,  and 
as  the  loading  is  unusual  it  will  not  be 
further  considered. 

At  point  31  of  arc,  for  loc.  loads  at  o,  1, 
2,  3,  tender  at  4  to  9  inclusive,  c  —  -  3I'i) 
H-  74(1  =  —  .  5  or  less  than  for  any  other 
point  hitherto  examined.  This  value  will 
doubtless  be  increased  slightly  by  moving 
live  load  one  or  two  divisions  to  the  left. 

Consider  next  the  right  springing  joint. 
M2  and  consequently  c2,  for  dead  load  is 
minus.  This  obtains  for  spans  of  about 
35  ft.  and  upwards,  for  rise  =  l/5  span,  so 
that  the  live  load  to  left  of  crown  giving  a 
positive  moment  at  K)1  acts  against  the 
dead  load;  hence  we  should  not  expect  to 
find  c2  for  such  spans  as  great  as  cx  at  the 
left  springing,  where  the  moments  due  to 
both  live  and  dead  loads  have  the  same 
(minus)  sign,  but  only  trial  can  determine. 
For  spans  less  than  about  35  ft.,  max.  c2, 
may  be  greater  than  max.  c13  as  in  fact  was 
found  to  be  the  case  for  an  arch  of  25  ft. 
span. 


133 

For  this  100  feet  span,  M2  -f-  H  was  com- 
puted for  the  front  of  the  loc.  load  at  2',  1', 
0  and  1  successively,  and  found  to  be  great- 
est when  the  loc.  loads  were  at  0,  1,  2,  3 
and  tender  loads  at  4,  5,  G,  7,  8,  1).  The 
computation  proceeds  as  before,  the  sum  of 
the  coefficients  in  columns  M2  and  H  of 
the  general  table,  for  the  points  above  be- 
ing multiplied  by  hW  and  W  respectively, 
thus : 

-f  .923  X  20  X  27.8  =  -f  513.2 
+  .853  X  20  x  11.1  =  +  189.4 
Dead  Load  M't  =  —  250.2 


Total,  M2  =  +  452,4 

4.382  X  27.8  =  121.8 
2.597  X  11.1  =    28.8 
Dead  load  thrust  =  594.9 


Total,  H  =  745.5 

c2  ==  +  452,4  -  745.5  =  +  .61 

The  computation  for  max.  ct  proceeds 
after  the  same  principle.  The  maximum 
GI  corresponds  to  loc.  loads  at  5,  0,  7,  8  and 
tender  load  at  9.  The  moment  coefficients 
are  taken  from  column  M,  in  general  table. 


134 

-  1.0648  X  20  X  27.8  =  —  592.0 

-   .1888  X  20  X  11.1  =-     41.9 

—  250.2 

Total,  M,  =    -  884.1 

1.6998  X  27.8  =  47.25 
.0497  X  11.1  =       .55 
504.90 

Total,         H  =  642.70 
ct  =    -884.1  -=-  642.7  =  — 1.370,  or  the  re- 
sultant  passes  1.38  ft.  below  the  centre  of 
the  left  springing  joint  for  this  position  of 
the  live  load. 

To  draw  the  left  resultant  in  position 
we  next  compute  the  vertical  component 
Va.  Add  up  the  coefficients  in  column  V,, 
of  general  table  for  points  5,  6,  7,  8  and 
multiply  by  27.8;  also  multiply  (for  tender 
load  at  point  9)  .9918  by  11.1;  the  sum 
added  to  the  half  weight  of  arch  gives  the 
total  Vx 

3.0295  X  27.8=  100.S 

.9918  X  11.1  =    11.0 

%  weight  arch  =  5 00. 7 

Total,  V,  =  078.5 


135 

From  the  point,  1.38  ft.  below  the  centre 
of  the  left  springing  joint,  lay  off  verti- 
cally upwards  V\  =  678.5,  to  any  scale,  and 
from  the  upper  extremity  of  this  line  draw 
a  horizontal  equal  in  length  (to  the  scale 
of  VJ  to  total  H  =  042.7  to  fix  the  pole 
of  the  force  diagram.  A  line  from  the 
pole  to  the  lower  extremity  of  Yt  gives 
the  magnitude  and  direction  of  the  result- 
ant on  the  left  springing  joint.  It  cuts 
this  joint  .95  ft.  below  its  centre  For  ac- 
curacy the  left  springing  joint  should  be 
drawn  to  a  large  scale  (anywhere  along 
the  radius),  and  this  construction  made  to 
the  large  scale  as  this  is  found  to  be  the 
joint  where  the  centre  of  pressure  is  far- 
thest from  the  centre. 

As  the  resultant  passes  0.13  ft.  outside 
of  the  middle  third  the  depth  of  key  must 
be  increased.  From  this  and  some  other 
examples  it  was  found  that  if  the  depth  of 
key  was  increased  by  3  to  4  times  the 
departure,  the  line  of  resistance  for  the  new 
arch  would  lie  inside  the  middle  third  lim- 
it, just  touching  it  at  the  critical  joint,  the 
left  springing  in  this  instance. 


In  this  case  the  depth  of  key  was  in- 
creased  by  4  X  0.13  =  0.52  ft.,  or  say  0.5 
ft.,  so  that  the  new  key  was  5.5  feet.  A 
new  construction  and  computation  for  this 
arch  showed  that  the  resultant  on  the  left 
springing  joint  passed  .02  ft.  inside  the 
lower  middle  third  limit  or  practically 
touched  it. 

If  desired,  the  equilibrium  polygon  for 
the  entire  arch  can  now  be  drawn  for  the 
last  loading  considered.  As  before  we  com- 
pute Vt,  H,  ct,  and  it  is  best  to  compute 
c2  as  a  check  on  the  construction  which  can 
be  made  as  explained  in  Art.  loa.  It  is 
much  better,  however,  to  find  by  computa- 
tion the  position  of  the  centre  of  gravity 
of  the  left  half  of  arch  and  load  c^nd  deter- 
mine by  construction  at  once  the  centre  of 
pressure  at  the  crown  joint,  from  which 
points  the  polygon  can  be  drawn  more  ac- 
curately towards  either  abutment. 

28.  On  investigating  arches  of  various 
spans  in  the  manner  indicated  above,  it 
was  found  that  the  springing  joints  were 
the  only  ones  necessary  to  examine,  except 
in  the  case  of  a  12.5  ft.  span,  2.5  ft.  rise 


137 

and  2.2  ft.  depth  of  key,  where  a  single 
concentrated  load  of  40,000  pounds  over 
point  6  was  found  to  cause  the  resultant 
on  the  joint  through  0  to  reach  the  upper 
middle  third  lin  it.  The  load  in  no  other 
position  gave  as  great  a  departure  on  any 
other  joint.  In  this  case  the  values  of  c, 
V,  and  H  were  computed,  and  from  the 
force  diagram  resulting  the  true  direction 
of  the  resultant  on  joint  0  in  position  and 
magnitude  was  obtained. 

In  any  arch,  the  resultant  on  the  critical 
joint  having  been  found  in  position  and 
magnitude,  the  normal  component  can  be 
scaled  off  and  the  maximum  intensity  of 
stress  at  the  most  compressed  edge  found 
as  in  Art.  21. 

The  following  table  gives  the  final  results 
of  a  series  of  constructions  to  determine 
the  depth  of  key  for  stone  arches  of  rise 
—  one-fifth  the  span,  so  that  the  line  of 
resistance  should  everywhere  be  contained 
within  the  middle  third  of  the  arch  ring 
of  uniform  cross- section  and  just  touch  it 
at  the  critical  joints. 

The  arch  ring  was  supposed  to  weigh 


138 

140  Ibs.  per  cubic  foot,  the  material  above 
it  112  Ibs.  per  cu.  ft.,  and  the  loading 
as  given  above,  viz. :  6,000  Ibs.  per  foot  of 
track  locomotive  load  for  about  20  feet 
(depending  on  the  length  of  the  horizontal 
divisions  of  the  arch),  followed  by  a  tender 
load  of  2,400  Ibs.  per  foot  on  about  30  feet, 
and  this  followed  by  a  second  locomotive 
and  tender  of  the  same  weights,  with  the 
one  exception  of  the  12.5  ft.  span,  where  a 
load  of  40,000  Ibs.  on  two  drivers  was 
alone  assumed.  These  loads  were  supposed 
to  bear  equally  on  8  feet  cross  ties  and  to 
be  transmitted  vertically  to  the  arch. 

The  actual  lengths  of  locomotive  and 
tender  loads  measured  along  the  rails  for 
the  different  spans  was  as  follows : 


Span. 

Length  loc.  load. 

Do.  tender  load. 

25, 

12.00, 

none, 

50, 

15.60, 

none, 

75, 

19.50, 

none, 

100, 

20.80, 

5.2, 

125, 

19.38, 

12.92, 

150, 

23.25, 

15.50. 

It  is  only  in  the  case  of  the  150  ft.  span 


139 


that  the  locomotive  length  appreciably 
exceeded  20  ft.,  but  the  weight  of  arch 
was  so  great,  compared  with  that  of  the 
load,  that  the  error  in  finding  the  depth 
of  key  was  small. 

TABLE  GIVING  THEORETICAL  DEPTH  OF  KEY. 


1 

Rise 

Key 

Loo  1  Load 
at 

Tender 
at 

Maximum 
Intensity 
of  Stress 

Feet 

Feet 

Feet 

Tons  pr.  eq.  ft. 

12.5 

2.5 

2.2 

6 

none 

25. 

5.0 

2.G 

0.1.  ...9 

«* 

9. 

50. 

10.0 

3.5 

4.  ...9 

" 

14. 

7n. 

15.0 

4.5 

5  ....  9 

*• 

22. 

100. 

20.0 

5.5 

5  ....  8 

9 

25. 

125. 

25.0 

6.25 

5.  6.  7 

8,9 

HO 

150. 

30.0 

7.                 5,  6.  7 

8.  9 

36. 

As  mentioned  above,  the  joint  of  rupture 
where  the  departure  of  the  resistance  line 
from  the  centre  of  joint  was  greatest,  was 
found  to  be  the  left  springing,  except  for 
the  25  ft.  span,  where  it  was  the  right 
springing  and  the  12.5  ft.  span,  where  6 
was  the  critical  joint. 

The  above  values  are  plotted  to  scale  ' 
Fig.  25,  being  shown  by  the  small  circles. 
A    line   through    these    circles   is   nearly 
straight,  but  not   sufficiently  so  for  accu- 
racy.     The   following   table   gives    these 


and  interpolated  values  for  every    5  feet 
for  use  of  constructors: 


DEPTH  Or  KEY  !N  FEZT 

~T 


o,|~Z 


141 

DEPTH  OF  KEY  FOB  ARCH  OF  UNIFORM  SECTION  AND  RISE 
—  \  SPAN,  ALL  DIMENSIONS  BEING  IN  FEET. 


Span 

Key 

Span 

Key 

Span 

Key 

5 

1.96 

55 

3.7 

110 

5.80 

10 

2.12 

60 

3.9 

115 

5.95 

12.5 

2.20 

65 

4.1 

120 

6.10 

15 

2.27 

70 

4.3 

125 

6.25 

20 

2.43 

76 

4.5 

130 

6.40 

25 

2.60 

80 

4.7 

135 

6.55 

30 

2.77 

85 

4.9 

140 

6.70 

35              2.95 

90 

5.1 

145 

6.85 

40             3.13 

95 

5.3 

150 

7.00 

45              3.30 

100 

5.5 

155 

7.15 

50               tt.5<) 

106 

5.65 

160 

7.30 

On  fig.  25  are  plotted  for  comparison  the 
depths  of  key  purposed  by  Dejardin  (line 
D),  Scheffler  (line  S,  by  interpolation  from 
his  tables  for  rise  =  V.4  and  */6  span),  Croi- 
zette-Demoyers  (line  C  D)  and  Trautwine 
(line  T)  (see  Art.  31). 

The  depths  of  key,  as  computed,  are  in 
excess  of  most  of  the  values  given,  all  of 
which  refer  to  materials  of  only  average 
strength  (second  class  masonry  for  the 
Trautwine  line).  This  excess  was  to  be 
expected,  for  most  of  the  old  formulas  are 
founded  on  the  successful  practice  of  the 
past,  and  cannot  therefore  be  expected  to 
give  results  corresponding  to  the  very  much 
heavier  locomotive  loads  of  to-day,  though 


142 

'some  of  them  may  be  a  rude  sort  of  a  guide 
in  the  design  of  common  road  bridges. 

The  French  authors  quoted,  after  find- 
ing  the  depth  of  key  (as  plotted  above), 
then  increase  the  radial  length  of  joint  to- 
wards the  abutment,  making  it  vary  as  the 
secant  of  the  inclination  to  the  vertical. 

The  above  table  is  for  an  avch  of  uni- 
form section.  The  same  material  can  be 
better  disposed,  perhaps,  by  making  the 
depth  of  keyless  and  increasing  the  length 
of  joint  as  we  approach  the  abutment;  but 
the  theoretical  treatment  of  this  case  falls 
under  that  of  the  arch  of  variable  cross- 
section,  and  it  has  to  be  omitted  for  want 
of  space. 

The  table  gives  the  depth  of  key  for  im- 
movable piers  or  abutments  and  arch 
stones  that  fit  perfectly  between  the  skew- 
backs,  when  laid  on  the  centres  (without 
mortar)  and  supposed  not  under  stress. 
Where  the  abutments  or  pier*  are  yielding^ 
either  from  not  having  a  rock  foundation 
or  from  too  small  a  width  (particularly 
in  a  series  of  arches),  or  where  mortar 
in  the  joints  is  used  which  is  not  hard 


143 

when  the  centres  are  struck,  or  in  any  case 
when  the  mortar  joints  are  not  thin  and 
hard,  so  that  the  arch  cannot  be  regarded 
as  practically  homogeneous  throughout, 
then  an  increase  should  be  given  to  the 
depths  above,  according  to  the  best  judg- 
ment of  the  engineer. 

The  effect  of  temperature  on  the   arch 

been    omitted  though    its    effects    are 

large    (see  Science   Series,   No.  48).     The 

•  lynamic  effect  of  the  load  is  supposed  to 

In'  taken  up  by  the  spandrels. 

The  arch  ring  has  been  supposed  above 
TO  be  of  sandstone  and  weigh  140  Ibs.  per 
ruble  foot.  If  it  weighed  100  Ibs.  the 
same  depth  of  key  would  correspond  to 
live  loads  8/7  of  those  assumed,  or  a  less 
depth  of  key  would  suffice  for  same  loads. 

^.).  MAXIMUM  INTENSITY    OF    STRESS    AL- 
LOWABLE  IN  STONE   AND   BRICK  ARCHES. 

The  average  pressure  on  a  joint  is  equal 
to  the  normal  thrust  divided  by  the  area 
of  the  joint,  and  this  reaches  a  maximum 
in  existing  bridges  (according  to  Scheffler, 


144 

for  masonry  weighing  150  Ibs.  per  cu.  ft.) 
from  17  tons  per  square  ft.  at  the  crown  to 
62  tons  per  square  ft.  at  the  springing.  In 
such  arches  the  resultants  on  the  joints  acl 
outside  the  middle  third;  but  even  if  they 
acted  at  the  middle  third  limits,  the  inten- 
sity of  pressure  at  the  most  compressed 
edge  would  be  double  the  above  or  .34  and 
124  tons  per  square  ft.  at  crown  and 
springing  respectively.  Scheffler  recom- 
mends not  exceeding  average  pressures  at 
the  crown  and  springing,  corresponding  to 
columns  of  the  same  material  as  the  vous- 
soirs  207  and  308  ft.  high,  or  say  15.5  to  *^> 
tons  per  square  ft.  for  stone  weighing  150 
Ibs.  per  cubic  ft. 

This  rule  seerns  safe  (for  cement  joints,  not 
common  mortar)  and  corresponds  for  stone 
weighing  150  Ibs.  per  cu.  ft.  or  fair  lime- 
stone, to  intensities  of  pressure  at  the  most 
compressed  edges  of  31  tons  per  sq.  ft.  at 
the  crown  and  40  tons  at  the  springing. 

30.  Existing  structures  that  have  done  good  service,  :IH 
well  as  arches  which  have  failed,  afford  the  data  from 
which  the  many  empirical  formulas  for  depth  of  keystone 
have  been  derived.  These  formulas  are  not  based  on  theory 
but  on  successful  practice  and  are  valuable  in  their  way, 


145 

but  very  unsatisfactory  iii  some  respects.  Thus  they  differ 
very  greatly  in  their  results,  some  giving  double  the  depth 
of  Keystone  for  certain  spans  as  others,  and  besides  they 
rarely  make  a  distinction  between  a  common  road  bridge 
and  one  intended  for  the  heaviest  modern  locomotives  to 
pass  over  at  great  speed. 

In  fact  these  formulas  generally  represent  the  practice 
of  the  past,  mainly  for  light  road  bridges  (with  a  few  excep- 
tions) and  can  serve  a  useful  purpose  in  the  design  of  such 
bridges;  but  most  of  them  are  evidently  inadequate  for  the 
heavy  moving  loads  of  to-day  on  railroads. 

As  preliminary  to  writing  some  of  the  best  known  of 
the  formulas,  it  will  be  convenient  to  give  simple  formulas 
for  expressing  the  radius  in  terms  of  the  span,  for  ease  of 
reduction,  as  most  of  the  formulas  are  expressed  in  terms 
of  the  radius. 

If  we  call  s  =span  in  feet,  h  =  rise  in  feet  of  a  circular 
arch,  r  being  the  radius  of  the  intrados,  we  have  the  w« -11- 
kuown  relative 


_      h       1  29 

For  —  = —  ,  r—  —  s 
s       5  40 


,    ••   -s 


I,     "    1.8  . 

io 


I'l  24 

31.  The  following  formulas,  by  French  engineers,  for  k  = 
depth  of  keystone  in  feet,  are  taken  from  DuBosque  ("Pouts 
et  Viaducts  en  Maconnerie")  after  reducing  to  English 


146 


equivalents.  They  are  intended  to  apply  to  best  bricks  or 
to  stone  not  so  hard  as  granite,  stone  of  "medium  resist- 
ance" as  BuBosque  styles  it. 

Perronefs  formula  is  intended  for  every  kind  of  arch, 
semi-circular,  segmental,  elliptical  or  basket  handle,  and  is 
as  follows  : 

k  =  l  4  .0347s  ....  (2). 

A  more  recent  author,  DC  jar  din,  gives  the  following  for 
circular  arches  : 

— =— ,  k  =  l  4  0.1  r  =  l       .'i.-, 


— =4-    k=l+.05r=l       J> 


_=_  ,k  =  14.035r=14  .037s 

s        s 

—=  —  ,  k  =  1  4  .02  r  =  1  4  .026  s   | 
s      10 

and  for  ei    ptical  or  basket  handled  arches 

— =i,k  =  14  .07  r  ....  (4) 
s       3 

in  which  r  equals  the  radius  of  curvature  at  the  crowrn. 

Another  French  authority,  M.  Croizette  Desnoyers,  gives 
the  following  for  segmenta  arches,  the  first  also  applying 
to  semi-circular  arches. 

v  —  t\  ^JL,  97  \/  •).  r 
6 


=  0.5  4. 27  \/  2  r. 

I 

, ,—     I 

,5  4  .253  V  '2  r   =.54  .327  V  s,     | 


.5 


.235\/2r   =  .54.342v/s,      \  ....(5) 


— =— ,  k  =.5  4. 217  v7  2  r   : 
s      10 


.5  4    .35  V/S, 


147 

He  likewise  adapts  the  first  formula  of  (5)  to  elliptical 
or  false  elliptical  arches  of  small  rise,  by  considering  r  to 
represent  the  radius  in  feet  of  an  arc  of  a  circle  of  same 
span  and  rise. 

Dupuit  gives  a  much  smaller  depth  of  key  by  the  fol- 
lowing formulas : 

— =  —  ,  k  =  o.:«j  v 

s       2 

h       1  >   ••"(fi) 

—  <— ,k  =0.27  v/  B=  V  .073  s    | 

J 

These  call  for  good  granite  laid  with  care. 

To  all  the  above  formulas  must  be  added  .02  //,  u-Jiere  H 
is  the  height  of  the  surcharge  above  the  crown,  reduced  if 
necessary  to  the  density  of  earth. 

This  is  plainly  a  very  ru  It:  and  i  u-x.kot  way  of  allowing 
for  an  extra  surcharge  of  earth,  M:r-  i  ;v  moleraie  amount, 
must  add  materially  to  the  slahilit'/,  the  io;;d  being  fixed  and 
symmetrical  with  respect  to  the  crown  and  thus  giving  a 
line  of  resistance  much  nearer  the  centre  line  of  the  arch 
ring  than  an  eccentric  rolling  load,  and  not  calling  for  an 
extra  section  on  account  of  the  dynamic  effect  of  the  live 
load.  A:tually  a  smaller  ar  /h  ring  can  be  used  up  to  a  * 
certain  height  with  the  same  security,  especially  consider- 
ing that  the  active  (or  passive)  pressure  of  the  earth  around 
the  arch  almost  ensures  stability  (when  crushing  is  not  to 
be  feared)  even  for  thin  arch  riu^s. 

In  any  case  the  arch  should  be  examined,  first  leaving 
out  the  horizontal  pressure  of  the  earth,  which  generally 
only  adds  to  the  stability,  and  afterwards  considering  it. 

After  finding  the  depth  of  keystone  by  preceding 
formulas,  the  Europeans  generally  increase  the  (radial) 
depth  of  arch  ring  from  the  crown  to  the  springing.  If 
we  call  d  the  angle  that  any  joint  makes  with  the  vertical, 
and  I  the  radial  length  of  any  joint  for  seymental  archcx, 
the  following  formula  for  this  length  is  frequently  used, 

1  =  k  sec  d (7). 

Du  Bosque  prefers  to  find  the  radial  length   of  the  joint 


148 

at  the  springing  joints  and  crown  by  (7)  and  other  formu- 
las above,  and  draw  an  arc  of  circle  through  the  upper 
ends  of  those  joints  to  define  the  extrados. 

For  semi-circular,  elliptical  or  basket  handled  arches 
the  rule  is  to  measure  up  from  the  springing  line,  half  the 
rise  and  draw  a  horizontal  to  intersection  with  the  intrados  ; 
the  joints  drawn  there  normal  to  the  intrados  called  "  the 
joints  of  rupture,"  must  have  a  length  equal  to  the  depth 
of  keystone  multiplied  by  a  coefficient,  which  is 

2    for  semi-circles 

1.8   "  ellipses,  &c.,  rise  =  —  span. 


1.4  «          "         «       -   =  1      .. 
5 

As  before,  a  circle  is  drawn  through  the  upper  ends  of 
the  "  joints  of  rupture,"  and  the  crown  joint  for  the  ex- 
trados down  to  the  joint  of  rupture  and  there  tangents  are 
drawn  to  this  circle  to  limit  the  masonry  down  to  the 
abutment.* 

•  The  above  lengths  of  joints  at  crown  and  elsewhere  are 
in  excess  of  average  English  and  American  practice,  which 
may  be  partly  due  to  the  poorer  qualities  of  stone  found  in 
France.  We  shall  now  give+some  English  and  American 
formulas. 

Rankings  formulas.  Find  the  longest  radius  of  curva- 
turer  of  the  arch  ;  then  the  depth  of  key  for  a  single  arch, 
including  tunnel  arches  in  rock  or  conglomerate,  is 

k=  y  0.12  r    •     •     •     •     (8)- 

For  an  arch  of  a  series  the  coefficient  of  r  under  the 
radical  should  be  0.17;  fora  tunnel  arch  in  gravel  or  firm 
earth  0.27,  and  in  wet  clay  or  quicksand  0.48.  The  defect 

*  See  Van  Nostrand's  Magazine  for  December,  1883,  for 
illustrations  in  article  by  E.  Sherman  Gould,  C.  ~E. 


149 

in  this  formula  is  the  lack  of  a  constant  term  to  give  a 
proper  depth  of  key  for  arches  of  small  span.  It  gives 
smaller  values  even  than  Dupuit's  formula  for  the  very 
best  materials. 

Trautwine's  formula,  for  first-class  masonry, 


k  =0.2  +  14  Vr-f0.5s    ....     (9) 

has  the  constant  term,  but  is  wrong  in  principle,  in  that 
for  the  same  span  the  depth  of  key  increases  with  r  or  as 
the  rise  diminishes  ;  whereas  the  flatter  the  arch,  for  the 
same  span,  the  less  should  be  the  key  (where  crushing  is  not 
in  question)  since  a  line  of  resistance  can  be  more  easily 
inscribed  within  the  same  limits  in  a  flat  arch  than  in  one 
of  greater  rise  when  the  key  is  the  same.  This  last  defect 
characterizes  all  the  formulas  given  above,  but  those  of 
Dejardin  and  Dupuit. 

Trautwine  increases  the  depth  given  above  >e  for  second 
class  masonry,  and  about  %  for  brick  on  fair  rubble. 


150 


CHAPTER  V. 

PRINCIPLES     AFFECTING     SOLID     ARCHES 
FIXED  AT  THE  ENDS.     LEMMA. 

SECOND   GENERAL   METHOD    OF   LOCATING 
THE  TRUE  LINE  OF  RESISTANCE. 


320  Fundamental  Equations  of  Solid 
Arches  "fixed at  the  ends" 

Space  forbids  deducing  the  fundamental 
equations  of  solid  arches,  but  the  reader  is 
referred  to  the  author's  "  Theory  of  Solid 
and  Braced  Elastic  Arches,"  pages  21  to 
3G,  for  a  simple  development  of  the  theory. 
The  following  are  the  three  equations 
which  must  be  satisfied  in  order  that  a 
line  of  resistance  may  be  the  true  one: 


El 


151 

(1), 


My     - 
KI     =0 


The  first  indicates  that  the  end  tangents 
to  the  centre  line  of  the  arch  ring  are  fixed 
in  direction;  the  second,  that  the  deflec- 
tion of  one  end  of  the  arch  below  the 
other  is  zero ;  and  the  third,  that  the  span 
is  invariable. 

The  centre  line  of  the  arch  ring  is  sup- 
posed divided  into  a  great  number  of  parts, 
each  equal  to  AS;  M  represents  the  mo- 
ment of  the  resultant  about  the  centre  of 
the  joint  traversing  the  middle  of  the 
corresponding  AS,  E  is  the  modulus  of 
elasticity  for  the  corresponding  voussoir 
AS  long,  and  I  represents  the  moment  of 
inertia  of  a  plane  joint  traversing  the 
centre  of  AS,  about  a  horizontal  axis 
passing  through  the  centre  of  the  joint. 
The  origin  of  co-ordinates  is  taken  at  the 
centre  of  the  left  springing  joint;  x  is  the 


152 

horizontal  and  y  the  vertical  distance  from 
this  origin  to  the  centre  of  the  correspond- 
ing AS.  The  summation  extends  over  the 
entire  arch  ring. 

If  we  call  H  the  uniform  horizontal 
thrust  of  the  arch  for  vertical  loading,  and 
v  the  vertical  distance  from  the  centre  of 
any  joint  traversing  the  middle  of  AS,  to 
the  resultant  acting  on  that  joint,  we  have 
by  Art.  22,  M  =  Hv. 

As  by  the  graphical  method  it  would  be 
impracticable  to  divide  the  centre  line  of 
the  arch  ring  into  a  very  great  number  of 
parts,  we  must  content  ourselves  with  divid- 
ing it  into  a  certain  number  of  parts  of 
appreciable  length  and  find  the  M,  E,  I,  x 
and  y  for  the  middle  of  each  part,  which 
gives  a  fairly  good  average  and  is  suffici- 
ently correct  in  practice. 

If  we  regard  the  modulus  E  as  constant 
throughout  the  arch  ring,  it  may  be 
droppe-i  from  the  equations. 

Similarly,  replacing  M  by  Hv,  H  may 
be  dropped  as  well  as  AS. 


153 

Therefore,  for  an  arch  ring  of  constant 
cross-section  where  I  is  constant,  the  three 
conditions  (1),  (2)  ard  (3),  reduce  very 
simply  to 

2(v)  =  0  ----  (4), 
2  (vx)  =  0  ....  (5), 
0  ....  (li). 


In  the  case  of  the  voussoir  arch,  if  the 
curve  of  the  centres  of  pressure,  as  deter- 
mined in  position  by  the  above  equations 
in  a  manner  to  be  shown,  lies  everywhere 
in  the  middle  third  of  the  arch  ring,  there 
will  be  no  tension  exerted  on  any  joint,  so 
that  the  theory  of  the  solid  arch  exactly 
applies  when  there  is  no  mortar  in  the 
joints  and  the  stones  are  cut  to  fit  perfectly. 

If,  however,  the  centre  of  pressure  on 
any  joint  without  mortar  lies  outside  the 
middle  third,  only  a  part  of  this  joint  is 
under  compression  (Art.  21),  so  that  on 
substituting  the  I  for  that  part  in  eqs.  (1), 
(-0  and  (o)  for  an  arch  of  variable  cross 
section,  a  nearer  approximation  can  be 
made  by  another  trial  and  so  on.  Event- 
ually the  assumed  and  computed  values  of 


154 

I  will  practically  agree  when  the  line  of 
resistance  can  be  regarded  as  fixed.  In 
case  there  is  mortar  in  the  joint  that  can 
supply  all  needed  tensile  resistance,  the 
line  of  resistance  can  pass  without  the  mid- 
dle third  without  any  change  in  the  equa- 
tions, as  the  voussoir  arch,  save  for  a  dif- 
ferent modulus  for  the  thin  mortar  joints, 
is  subjected  to  the  same  deformation  as  the 
corresponding  solid  arch,  so  that  its  line  of 
resistance  is  nearly  identical.  As  in  well 
designed  bridges  the  line  of  resistance 
should  nowhere  pass  out  of  the  middle 
third,  for  any  loading,  to  avoid  the  possi- 
bility of  joints  opening  with  the  accom- 
panying infiltration  of  water,  as  well  as  to 
provide  a  factor  of  safety,  the  tentative 
method  above  will  rarely  be  needed,  and 
the  line  of  resistance  can  be  at  once  found 
from  (4),  (5J  and  ((>)  for  the  arch  of  con- 
stant cross  section. 

33.    LEMMA. 

In  the  constructions  needed  for  establish- 
ing the  true  line  of  resistance,  according  to 


the  theory  of  the  solid  arch,  we  shall  have 
to  solve  the  following  problem : 


In  fig.  20,  having  given  a  series  of  fixed 
points  bT,  b2,  b3,  ...,  1>8,  and  having  drawn 
parallel  ordinates  1>3  v3,  1>4  v4,  ...,  through 
them,  intersecting  a  line  m,  m  in  the  points 
mT,  ni8,  ...,  it  is  required  to  draw  in,  m8  so 
as  to  satisfy  the  two  conditions. 

2  (m  b)  =  o,  ~  (mb  .  x)  —  o 

where  m  b  represents  the  ordinate  from 
ml  mg  to  any  one  of  the  points  b,  ordinates 
above  mt  m8  being  counted  positive,  those 
below  negative,  and  x  represents  the  dis- 
tance (abscissa)  from  an  assumed  origin  O 


156 

to  the  ordinate  m  b  to  which  it  refers.  If 
we  give  x  the  subscript  of  the  ordinate  to 
which  it  refers,  then  the  above  conditions 
can  be  written  in  full,  for  this  figure, 

m8  b3+  m4  b4  +  m5  b5  +  me  b6—  (m1  b,  +  m2 
b,4-m7b7+ni8b8)=o; 


(m8b8.x8+m4b4.x£:f:  m5  b5  .  x5  +  m6  be  . 

Xg)  —  (nij  bj  .  xx  +  m2  b2  .  x2  +  m7  b7  .  x7  + 

m,  bs  .  xs)  =  o. 

Now  draw  a  straight  line  from  b1  (= 
vj  to  bs  (=  v8)  and  designate  where  it  in- 
tersects the  ordinates  by  v2,  v3,  .  .  .  ;  any  or- 
dinates  as  m5  b5  can  be  written 


or  generally, 

mb  —  vb  —  vm  ; 

which  gives  m  b  plus  when  above  mx  ms 
and  minus  below,  as  is  imperative.  Sub- 
stituting in  the  equations  above  we  have 

2  (vb  —  -vm)  =  o,  2  (vb  —  vm)  x  =  o. 
.-.  2(Jb)  =  2  (vm),  2(v¥)  .  x)  =  2  (vm  .  x), 


157 

If  we  call  x0  the  abscissa  of  the  result- 
ant of  the  lines  of  the  type  vb  treated  as 
forces  and  x'0  the  abscissa  of  the  resultant 
of  the  lines  vm  treated  as  forces,  then,  since 
the  moment  of  the  resultant  is  equal  to  the 
sum  of  the  moments  of  the  components,  the 
last  eq.  above  can  be  written, 

x0  2  (vb)  =  x'u  I;  .Tin) ; 

whence,  in  view  of  the  relation,  3  (vb)  = 
2  (vm),  we  have  x0'  =  x0.  This  establishes 
the  proposition,  that  when  the  line  n^  m8 
has  been  determined  correctly,  the  result- 
ant R  of  the  lines  vb,  treated  as  forces, 
must  equal  and  coincide  with  the  resultant 
of  the  lines  vm  treated  as  forces. 

We  can  quickly,  to  any  convenient  scale, 
find  the  value  of  R  =  v2  b,  +V3  m~  •  -  • 
+v7  b7.  Its  position  can  be  found  by  tak- 
ing moments,  most  conveniently,  about  an 
ordinate  AB  through  the  centre  of  the  liiu- 

b,  b, 

If  the  lines  v  b,  by  pairs,  are  equidistant 


158 

from  AB,  as  happens  in  all  the  applica- 
tions that  follow,  call  the  distances  from 
this  ( dotted )  medial  ordinate  ( AB)  to 
the  ordinate  through  bj?  b2,  b3  and  b4, 
z,,  z2,  z3  and  z4  respectively.  Then 
treating  left  handed  moments  as  positive, 
right  handed  as  negative,  we  have  the  al- 
gebraic sum  of  the  moments  about  the  me- 
dial ordinate,  equal  to  (  v7  b7 — v2  b2 )  z2  + 
( v6  be— v8  .bs )  z3  -f(v5  b5  —  v4  b4 )  z4 ;  and 
on  dividing  this  by  R  (as  found  above)  we 
have  the  distance  from  the  medial  or- 
dinate to  R  which  can  then  be  laid  off  in 
position,  as  shown  in  the  figure. 

The  differences  as  (v7  b7  —  v2  ba)  can 
readily  be  found  by  taking  the  distance  v2 
b2  in  dividers  and  laying  it  off  from  v7 
along  v7  b7.  The  difference  between  the 
two  lines  is  to  be  measured  to  the  same 
scale  as  the  ordinates  v  b  in  finding  the 
value  of  R  above.  (This  method  is  to  be 
generally  used  in  similar  cases). 

We  next  draw  a  trial  line  nl  ng  and  di- 
vide ordinates  as  v5  n.  (n5  being  the  inter- 
section of  nj  ns  with  the  ordinate  v5  b5)  in- 


159 

to  two  sets  by  a  line  drawn  from  vt  (=bt  ) 

to  n8. 

The  resultant  T  of  the  sum  of  the  ordi- 
nates  from  the  line  v,  v8  to  v1  ng  can  be 
found  in  magnitude,  by  adding  up  the  or- 
dinates, and  in  position  by  taking  mo- 
ments about  A  as  just  explained.  Lay  it 
off  the  computed  distance  to  the  left 
of  A. 

Now  the  position  of  T  is  not  changed 
when  YJ  n8  assumes  its  true  position  v,  m8 
(m1  m8  being  regarded  as  the  true  line  to 
satisfy  the  original  conditions),  since  all 
the  ordinates  in  the  triangular  space  vt  vg 
ns  are  altered  in  the  same  ratio.  T  is  thus 
fixed  in  position  no  matter  where  n8  may 
be  placed  on  the  line  vs  n8. 

It  follows,  because  of  this  property 
and  since  the  ordinates,  by  pairs,  are 
equidistant  from  AB,  that  the  resultant 
T'  of  the  ordinates  intercepted  between 
Vj  n 8  and  nt  n8  is  at  the  same  distance  to 
the  right  of  A  that  T  is  to  the  left.  Then 
if  ul  is  afterwards  shifted  to  m19  T'  is  un- 
changed in  position,  since  all  ordinates  are 
altered  in  the  same  ratio.  Finally,  if  n8  is 


160 

shifted  to  m8,  m,  remaining  stationary,  the 
position  and  value  of  T"  remain  un- 
changed. Hence  lay  off  T'  in  position  as 
far  to  the  right  of  A  as  T  is  to  the  left,  and 
get  its  trial  value  by  adding  up  the  ordi- 
nates  included  between  vl  ns  and  n: 

iV 

From  what  has  been  proved  above  it  is 
plain  that  if  n:  n8  has  been  drawn  correct- 
ly, the  resultant  of  T  and  T'  or  of  the  or- 
dinates  between  v1  vg  and  nt  n8  must  coin- 
cide with  and  be  equal  to  R;  hence  calling 
I  and  I1  the  distances  from  T  and  T' 
respectively  to  R,  we  have 


If  the  trial,  T  representing  the  sum  of 
the  ordinates  from  vx  v8  to  v,  ng,  is  not 
equal  to  the  true  value  of  T  just  found,  re- 
duce the  distance  vg  n  to  vg  m8  in  the  ratio 
of  the  true  T  to  the  trial  T  just 
found. 

Change  v1  u1  to  v:  m1  in  the  ratio  of  the 
true  T'  (given  by  formula  above)  to  trial 
T'  =  sum  of  ordinates  from  vl  n,,  to  nl  na. 


161 

The  reduction  in  both  cases  is  best 
effected  graphically. 

As  the  sum  of  similar  ordinates  in  the 
triangle  vl  n8  ml  is  the  same  as  for  the  tri- 
angle Y!  ms  nij  and  their  resultant  has  the 
same  position,  it  is  evident  that  m,  m8  is 
the  true  closing  line  (as  it  is  called)  to  sat- 
isfy the  conditions. 

2  (mb)  =  o,  2  (mb  .  x)  =  o. 

It  is  easy  to  see  if  the  first  condition  is 
fulfilled  by  taking  the  successive  lengths, 
m3  bs,  m4  b4,  m5  b5,  m6  b6  in  dividers  and 
adding  up  along  a  straight  line.  Similarly 
add  the  lengths,  m,  b,,  m2  b2,  m7  b7,  m8  b8, 
along  the  same  straight  line.  If  the  two  to- 
tal lengths  agree,  the  condition  2  (mb)  =  o 
is  satisfied. 

When  the  ordinates  vb  are  equal  at 
equal  distances  from  the  medial  line  AB, 
R  must  coincide  with  AB.  Now  the  re- 
sultant of  2  (vm)  cannot  pass  through  the 
centre  unless  ml  m8  is  drawn  parallel  to 
v,  v8,  in  which  case  the  lines  vm  will  all  be 
of  equal  length  throughout.  Their  num- 
ber, in  the  present  instance,  is  8,  so  that  by 


162 

the  first  condition    2  (vb)  —  2  (vn 
have 


v,m1  =  vgm 

8 
which  at  once  determines  the  line  m,  nig. 

It  is  equally  correct  and  shorter  to  take 
the  sum  of  the  ordinates  vb  to  one  side  of 
AB  and  divide  by  4  when  the  total  num- 
ber of  ordinates  is  8. 

34.  We  shall  now  proceed  to  design  a 
series  of  stone  (or  brick)  arch  bridges,  whose 
rise  is  one-fifth  the  span^  so  that  the  line  of 
resistance  for  the  position  of  the  rolling 
load  tried  shall  just  be  contained  within 
the  middle  third  limits  of  the  arch  ring, 
and  the  intensity  of  pressure  on  any  edge 
of  a  voussoire  joint  shall  not  exceed  say  oO 
tons  per  square  foot  for  the  best  brick,  or 
50  tons  for  good  granite  or  sandstone. 

The  live  load  assumed  is  known  in  Coop- 
er's Specifications  as  c<  Class  extra  heary 
A".  We  give  below  the  distances  in  feet 
from  the  front  pilot  wheel  to  each  pair  of 
wheels  in  turn  and  on  the  same  line,  the 
weight  of  the  pair  of  wheels  in  tons  of  2000 
pounds  : 


L63 


Pair  of  Pilot  Wheels—  0         f net—  8  tons. 


"    Driver 

'      —  8-! 

"    —15 

" 

•       —13.83 

•'    —15 

« 

«       —18.33 

"    —15 

" 

'       —22  H3 

"    —15 

"    Tender 

<       —  20.92 

"    —  9 

« 

*       -40.42 

"    -  9 

<t          <« 

'       —45.25 

"    —  9 

"    Pilot 

—54.25 

"    —  8    ' 

The  position  of  the  pilot  wheel  of  the  sec- 
ond locomotive  is  given  last,  from  which  all 
the  wheels  of  the  second  locomotive  can 
be  located  when  desired.  For  short  spans 
the  above  load  may  be  used  or  40  tons 
equally  distributed  upon  two  pairs  of  dri- 
vers, seven  feet  centre  to  centre,  whichever 
produces  the  most  hurtful  effect. 

The  arch  ring  will  be  supposed  of  uni- 
form section  throughout  and  composed  of 
brick  or  sandstone  weighing  140  Ibs.  (.(/! 
tons)  per  cubic  foot.  The  material  above 
the  arch  ring  up  to  the  level  of  the  road- 
way will  be  supposed  to  have  a  specific 
gravity  eight-tenths  of  that  of  the  arch 
ring. 

The  load  on  a  pair  of  wheels  is  carried 
through  the  rails  and  cross- ties  to  the  bal- 


164 

last.  If  these  cross-ties  are  8  feet  in  length 
we  shall  suppose  the  load  uniformly  dis- 
tributed along  this  length,  so  that  for  one 
foot  length  of  cross  tie  the  load  will  be  but 
one-eighth  the  load  on  the  whole  length  of 
the  cross  tie.  The  load  for  one  foot  in 
length  of  tie  is  then  reduced  to  cubic  feet 
of  masonry  by  dividing  by  0.07. 

35.  Example  ./(see  plate,  fig.  27).  The 
figure  represents  an  arch  of  12.5  ft.  span, 
2.5  rise,  2.2  ft.  depth  of  keystone,  and  ra- 
dius 9.06  ft.,  with  a  surcharge  rising  2  ft. 
above  the  crown  to  the  level  roadway  and 
loaded  3  ft.  to  the  left  of  the  crown  with 
20  tons  on  8  ft.  cross  ties,  equivalent  to 
35.7  cubic  ft.  of  masonry  of  the  same  spe- 
cific gravity  as  the  arch  ring  (0.7  ton  per 
cubic  foot)  on  1  ft.  length.  The  dotted 
medial  lines  of  the  trapezoids  are  0.<v.  of 
the  same  lines  extended  to  the  roadway, 
thus  giving  the  "reduced  contour"  shown. 
It  is  evident,  for  this  small  span,  that  the 
alternative  load  of  80,000  pounds  equally 
distributed  on  the  two  pairs  of  drivers  7 
feet  apart,  must  produce  the  most  hurtful 
effect.  As  only  one  pair  of  drivers  can  get 


165 

on  the  half  arch  they  are  placed  as  shown, 
not  quite  */4  8Pan  to  left  of  crown.  The 
centre  line  of  the  arch  ring  is  divided  into 
'M  equal  parts,  the  dotted  joints  are  drawn 
as  shown  (only  a  few  of  these  are  drawn 
near  the  left  springing),  two  divisions  apart 
and  radial  lines  (full  lines)  are  drawn  mid- 
way between  them.  The  portion  of  the 
arch  ring  between  a  springing  joint  and 
the  iirst  clotted  joint  constitutes  one  artifi- 
cial voussoir,  and  the  portion  between  any 
two  consecutive  dotted  joints  likewise  con- 
stitutes a  voussoir. 

As  the  theory  of  the  solid  arch  requires 
that  we  find  the  moment  for  each  voussoir 
about  the  middle  of  its  centre  line,  or  where 
the  full  radial  lines  cross  it,  at  the  points 
al5  a2,  ...  ,  the  construction  of  Art.  15  ap- 
plies, or  we  take  the  division  of  the  arch 
included  between  the  full  radial  lines  in 
tabulating  quantities,  and  find  the  result- 
ants acting  on  these  full  line  joints  as  in 
Article  15. 

The  condensed  tables  of  loads  and  arms 
(made  out  as  explained  in  Art.  15)  are  as 
follows: 


166 

LEFT  SIDE. 


Joint      8 

7 
5.7 

6 

5 

4 

3 

2 

1 

0 

vS 

1.9 

9.7 

49.5 

53.9 

58.6 

63.4 

68.5 

71.2 

c 

.24 

.76 

1.26 

2.66 

2.77 

2.94 

3.15 

3  .  42 

3.55 

RIGHT  SIDE. 


Joint 

9 

10 

11 

12 

13 

14 

15 

16 

S 
C 

1.9 

.24 

5.7 

9.7 

13.8 

18,2 

22.9 

2.82 

27.7 

32.8 

35.5 

.76 

1.26 

3.78 

2.31 

3.36 

.3.87 

4.13 

The  loads  S  are  laid  off  on  either  side  of 
the  arch,  the  arms  either  side  of  the  crown, 
and  for  an  assumed  thrust,  shown  by  the 
upper  inclined  line  through  the  crown,  the 
various  resultants  on  the  joints  1  to  L6  are 
found  as  in  Art.  15.  These  resultants  in- 
tersect the  verticals  through  aly  a2,  .  .  .,  a,G 
at  the  points  b1?  b2,  .  .  .  ,  b16. 

36.  To  locate  the  line  k  k',  measure  the 
ordinates  from  a  straight  line   joining  at 
and  a16,  to  the  points  a1?  a2  .  .  .,.a8;  add,  di- 
vide by  8  and  lay  off  the  distance  verti- 
cally from  aj  to  k  and  from  ^ ,  to  k'  and 
draw  the  line  k  k'. 

37.  Regard  kk1  as  the  trial  closing  line 
for  points  b;  connect  ^  and  b16,  also  k  and 


167 

b16  by  straight  lines  and  find  R  and  the 
trial  T  and  T',  exactly  as  explained  in  Art. 
33,  in  position  and  magnitude.  The  posi- 
tions are  given  on  the  figure.  Trial  T  = 
23.35,  trial  T'=  14.1,  R  =  38.78; 

.-.  true  T  =  38.78 14|  =  22.81; 
5.22 

true  T'  =  38.78  f^|  =  15.97. 
5.22 


( 

Hence  we  must  lay  off  bx  m  =  b,  k  ,^~r- 

»,•).•)•> 

and  b16  m'  =  b16  k/  u'j  '• 

This  is  best  done  by  the  ratio  lines  as 
shown,  or  the  distances  may  be  computed 
and  laid  off  to  scale.  The  line  mm'  is 
thus  the  true  closing  line  of  Art.  33,  for 
points  bt  b:,  .  .  .  blti. 

3S.  The  ordinates  y,,  y2,  .  .  .,  y8  (from  a 
a17  toa15  a,,  .  .  .,  aj  are  next  scaled  off;  also 
the  ordinates  from  k  k'  to  a^  a,,  .  .  .,  a., 
called  k  a15  ka.:,  .  .  .,  k  a8,  and  we  find  the 
value  of  2  (k  a.y)  —  2  (k  a4  .  y4  +  k  a5  .  y. 
+  k  a, .  y6  +  k  a7 .  y7  4-  k  a  s .  y8  —  k  ax .  y!  - 
ka,  .  y,  — ka8y.)  =  11.50. 


168 

Next,  the  ordinates  from  m  m'  to  the 
points  b:,  b,,  .  .  .,  b16,  called  mb^  mb2,  .  .  ., 
in  b16  are  scaled  off  and  2  (nib  .  y)  is  found. 
In  the  present  instance  the  complete  ex- 
pression for  this  is  (m  b8  +  m  b9)  ys  +  (m  b7 
+  m  b10)  y7  +  (m  be  +  m  bn)  y6  +  (m  b.  + 
m  bJ2)  y  5  +  (mb4  +  mbi3)  y4  —  (ml>8  +  mbl4) 
y3  —  (m  b2  +  in  b15)  y2  —  (m  bt  -f  m  bl6)  jl 
=  lX.*-v;S,  ordinates  above  mm'  being  treat- 
ed as  positive,  those  below  negative. 

We  have  now  only  to  reduce  the  ordi- 
nates nib  in  the  ratio  of  11.50  to  lJS.^8  (by 
the  proper  ratio  lines),  and  lay  off  the  re- 
duced lengths. from  the  line  kk'  vertically 
up  or  down,  according  to  the  sign  of  m  b 
to  find  all  the  points  c,,  c,,  .  .  .,  clG  in  the 
true  equilibrium  polygon  for  the  arch.  The 
ordinate  from  m  in'  to  the  point  where  the 
trial  thrust  meets  the  crown  is  likewise  re- 
duced in  the  same  ratio  and  laid  off  from 
k  k'  to  fix  the  true  centre  of  pressure  on 
the  crown  joint,  .05  ft.  below  the  centre  of 
the  joint. 

The  reader  familiar  with  Prof.  H.  T. 
Eddy's  "  Constructions  in  Graphical 
Statics  "  will  recognize  that  the  above  pro- 


169 

cedure  is  founded  upon  his  beautiful  con- 
structions for  the  solid  arch  fixed  at  the 
ends. 

30.  It  will  now  be  shown  that  the  points 
<?,  located  as  above,  are  points  in  the  equi- 
librium polygon,  directly  over  the  centre 
of  the  artificial  voussoirs,  which  satisfy  the 
three  conditions  for  an  arch  fixed  at  the 
ends  (Art.  32) 
>•  (ac)  =  o,  2  (ac  .  x)  =  o,  2  (ac  .  y).  =  o. 

Referring  to  Art.  3*5  it  is  seen  that  the 
line  k  k'  was  located  in  a  manner  satisfy- 
ing the  conditions, 

2  (k  a)  =  o,  2  (k  a  .  x)  —  o  .  .  .  (A), 

as  shown  in  Art.  33. 

Lines  of  the  type  k  a  in  these  formulas 
refer  to  vertical  ordinates  measured  from 
kk'  to  at,  a2,  ...  al6.  Similarly  mb  repre- 
sents a  vertical  ordinate  from  line  m  m'  to 
bj  or  bL,  etc.;  ordinates  above  kk'  or  mm1 
being  regarded  as  plus,  those  below  minus. 

By  the  method  used  in  Art.  37  or  Art, 
33  the  line  m  m1  was  located  to  satisfy  the 
conditions, 

2  (mb)  =  o,  2  (mb  .x)  =  o. 


170 

The  ordinates  m  b  were  DOW  all  changed 
in  the  same  ratio,  which  does  not  affect  the 
position  of  mm1.  The  altered  ordinates 
were  next  laid  off  from  k  k',  the  new  value 
of  in  b  being  equal  to  k  c  in  the  figure. 

The  conditions  just  given  are  thus  satis- 
fied by  the  k  c'  s, 

/.  -2tkc)  =o,  ^(kc.x)  =  o...(B). 

Also  by  the  construction  of  Art.  08,  since 
every  m  b  has  been  altered  in  the  ratio 
11.50  to  18.28  to  the  corresponding  k  c,  2 


as  we  see  by  reference  to  the  equations  of 
Art.  38. 

If  the  right  member  of  the  last  equation 
is  transferred  to  the  left  member,  since,  kc 
-  k  a  =  a  c,  we  have,  2  (a  c  .  y)  =  o.  On 
subtracting  eqs.  (A)  from  (B)  and  writing 
the  equation  just  found  in  the  group,  we 
have 

2  (a  c)  =  o,  2  (a  c  .  x)  =  o,  2  (a  c  .  y), 

or  the  points  c  satisfy  the  conditions  for  an 
arch  "  fixed  at  the  ends." 


171 

40.  As  the  closing  line  m  m'  has  been 
shifted  to  kk'  and  the  ordinates  m  b  al- 
tered in  the  ratio  11.50  to  18.28,  by  the 
theory  of  equilibrium  polygons,  we  draw 
from  the  old  pole  O  (on  the  left)  a  parallel 
to  m  m1  (in  its  first  position)  to  intersection 
J  with  the  load  line,  then  a  horizontal  to 
the  right  a  distance  =  old  pole  distance  X 

lo.ivO  -r-»      i  •    «  /»       i  ic 

-  to  r  the  position  of  the  true  pole  for 
11. oO 

the  left  force  diagram.  This  is  easily  ef- 
fected graphically  by  laying  oif  J  L  and  J 
M  in  the  ratio  of  11.5  to  18.28  and  draw- 
ing M  P  parallel  to  L  I,  I  being  the  point 
where  a  vertical  through  O  intersects  the 
horizontal  through  J. 

The  new  pole  may  likewise  be  found  by 
the  method  of  Art.  14,  by  drawing  through 
O  a  parallel  to  a  line  connecting  bt  and  bl6 
to  intersection  with  load  line,  then  from 
this  point,  a  parallel  to  a  line  connecting 
the  points  Ci  and  cl6,  previously  found,  a 
distance  to  the  right  whose  horizontal  pro- 

18.28 
lection  =  old  pole  distance  X  7 

11.  oO. 

To  fix  the  new  pole   P'  on  the  right, 


172 

draw  PC"  equal  and  parallel  to  ray  P  C 
on  left. 

The  points  c  can  be  tested  by  drawing 
the  new  resultants  on  the  joints,  having 
given  the  new  poles  and  the  position  of 
the  thrust  at  the  crown.  These  resultants 
produced  to  intersection  with  the  respective 
joints  from  a0  to  al7  give  the  centres  of 
pressure  on  the  corresponding  joints. 

The  centres  of  pressure  all  lie  within  the 
middle  third  of  the  arch  ring,  except  at 
joint  G  where  the  thrust  passes  exactly  l/Q 
depth  from  centre.  The  intensity  at  the 
upper  edge  of  joint  0  is  therefore  double 
the  mean.  The  normal  component  of  this 
thrust  is  the  weight  of  03.2  cu.  ft.  of  stone 
=  03.2  X  .07  =  4.4  tons;  the  mean  pres- 
sure is  thus  4.4  -r-  depth  joint  '2.2  ft.  =  2 
tons  and  the  intensity  at  upper  edge  is 
therefore  4  tons  per  square  foot. 

41.  To  test  the  accuracy  with  which  the 
work  has  been  done,  we  measure  a  c  at  the 
points  ax  to  al6,  counting  distances  above  a 
plus,  below  minus;  then  find  the  co-ordi- 
nates x,  y,  of  each  point  &l  to  a16  regarding 
a0  as  the  origin,  x  being  measured  horizon- 


173 


tally  (along  a0  a17)  to  the  ordinate  through 
any  point  a  and  y  vertically  above  a0  a1T 
to  point  a  as  previously  stated,  and  finally 
from  the  products  as  shown  in  the  table: 


Joint 

ac 

y                 ac  .  y 

x              ac.x 

4 

-     .13 

1.89 

.246 

2.78                 .36 

5 

-|-  .40 

2.26 

.904 

3.69              1.48 

6 

2.50 

.950 

4.60 

1.75 

7 

2.70 

.486 

5.57              1.03 

8 

+   .06 

2  .  80 

.168 

6.52                .39 

15 

.92 

.074 

12.90 

1.03 

16 

-f   .30 

.32 

J'l'G 

13.66             4.09 

{-1.53 

+2.924 

J-10.13 

1 

—   .27 

.32                   .086 

.34                 .92 

2 

—  .21 

.92 

.193 

1.12 

.23 

3 

—   .07 

1.45 

.101 

1.92 

.13 

9 

-   .10 

2.80 

.280 

7.50 

.75 

1<) 

-   .18 

'1.  To 

.486 

8.48              1.53 

11 

—  .20 

2.50 

.500 

9.41 

1.88 

12 

-  .18 

2.  -26 

.406 

10.34 

1.86 

13 

-  .13 

1.89 

.246 

11.23 

1.46 

14 

—   .00 

1.45 

.130 

12.10 

1.09 

-1.43                              —2.428                            —9.85 

We  have  here  the  ratios, 


(—  ac) 


~  1.43  '  2  ( 

(+  ac  .  x) 


ac  .  y) 
10.13 


2.924, 

2.428  ' 


2  (—  ac.x)  ~     9.85' 
These  ratios  should  strictly  be  unity,  so 
that  the  conditions  of  an  arch  "fixed  at 
the  ends," 


174 

2  (ac)  =  o,  2  (ac .  y)  =  o,  2  (ac .  x)  =  o, 
may  be  satisfied.  The  results,  however, 
are  very  close,  as  will  be  apparent  on  sup- 
posing the  points  c  all  to  be  lowered  one 
hundredth  of  a  foot  only,  when  2  (+  a  c) 
will  become  -f-  1.47  and  2  ( —  ac),  —  1.  ")*^, 
the  minus  sums  now  being  the  greater.  The 
changes  will  evidently  be  equally  great  in 
the  other  sums,  so  that  we  have  accidentally 
here  determined  the  centres  of  resistance 
within  about  .01  foot,  even  on  this  small 
scale.  We  may  readily  rest  content  though^ 
with  half  a  tenth  of  a  foot  error  on  each 
joint  owing  to  unavoidable  errors  of  con- 
struction. See  the  next  example  where 
these  errors  are  as  pronounced  as  for  any 
arch  examined  and  yet  a  shifting  of  the 
points  c  by  half  a  tenth  of  a  foot  is  about 
all  that  is  necessary  to  satisfy  the  condi- 
tions. 

We  conclude  for  the  arch  just  examined, 
for  the  given  position  of  the  live  load,  that 
the  line  of  resistence  just  touches  the  mid- 
dle third  limit  at  one  point  only,  and  that 
it  possesses  the  proper  margin  of  safety 
both  as  to  strength  and  stability. 


175 

42.  Example  II.  A  segmental  stone  arch  of  25  ft.  span,  5 
ft.  rise,  radius  18.12  ft.  and  uniform  depth  of  arch  ring  2.5 
ft.,  was  next  examined. 

The  depth  of  spandrel  filling  over  the  crown  was  2  ft. 
and  the  live  i'oad-consisted  of  the  last  three  driving  wheels 
of  the  locomotive  above  specified  on  the  left  half  of  the 
arch,  no  load  on  right  half. 

The  loads  are  given  precisely  as  follows  :  15  tons  3.8  ft. 
from  crown,  15  tons  8.3  ft.  and  15  tons  12.8  ft.  from  crown. 

The  division  of  arch  ring  and  the  construction  gener- 
ally was  exactly  like  that  just  given  for  Example  I,  so  that 
it  is  not  necessary  to  enter  into  it.  The  true  thrust  at  the 
crown,  after  the  theory  of  the  solid  arch,  was  found  to  act 
.06  ft.  below  the  centre  of  the  crown  joint  and  its  horizontal 
component  was  120.25  cu.  ft.  =  8.4175  tons,  the  vertical  com- 
ponent 10'.  5  cu.  ft.  =0.735  ton. 

If  we  call  for  brevity,  the  distance  ac=v,  we  find  in 
this  instance, 


_2_(_jv)  _  UK).     2  (+vx)  _25.64       2  (+vy}_  _6.32_ 
"YPv)  ""^53  '     2  (—  vx)  "18^4  '    3T(^-~vy)~  "BUT 

If  we  conceive  the  thrust  at  the  crown  lowered  0.1  ft. 
but  maintaining  its  same  direction  and  magnitude,  the 
points  c  will  all  be  lowered  0.1  foot,  and  the  new  ratios  will 
be  as  follows  : 

_?  lr.l>  _  126    s  (+vx).  _  16-^8     2(4-vy)  _  -2.96 

2  (—v)  ~2.30  '    "S  (—  vx)  ~~  28~75  '     2  (—  vy)  ~~  8.02 

Here  the  minus  t-rnis  exceed  the  plus  terms  so  much 
that  it  is  plain  that  the  thrust  has  been  lowered  too  much  ; 
in  fact  (neglecting  any  possible  tilting)  it  is  evidentsthat  the 
true  position  of  points  c  lies  between  the  first  and  last  posi- 
tions and  is  nearer  the  former  than  the  latter;  so  that  we 
can  safely  say  that  the  first  series  of  points  is  certainly 
within  0.05  foot  of  the  correct  position. 

As  the  discrepancy  above  was  as  great  as  that  found  in 
any  case  examined,  it  was  thought  worth  while  to  show  that 


176 

the  extreme  limit  of  error  in  any  rase  was  negligence  and 
that  the  construction  affords  practically  exact  results. 

On  constructing  the  centres  of  pressure  on  all  the 
joints,  it  was  found  that  they  nowhere  leave  the  middle 
third  except  at  the  right  springing  joint  (17)  where  the  cen- 
tre of  pressure  was  0.08  ft.  above  the  middle  third  limit. 
Hence  it  was  thought  best  to  increase  the  depth  of  keystone 
three  times  this  amount,  or  0.25  ft.,  making  the  radial  depth 
of  arch  ring  uniformly  2.75  f*et. 

This  is  so  near  the  former  value,  2.5  ft.,  that  it  was  not 
thought  worth  while  to  test  it  by  another  construction. 
It  will  be  assumed  to  satisfy  all  conditions. 

The  intensity  of  thrust  at  the  upper  edge  of  the  right 
springing  joint  (for  the  arch  ring  2.5  ft.  deep)  is  found  to  be 
9.2  tons  per  square  foot,  at  the  lower  edge  of  the  left  spring- 
ing joint  8.7  tons. 

43.  Example  III.  Segmental  stone  arch  of  50  feet  span, 
10  ft.  rise,  radius  36.25  ft.,  and  height  of  surcharge  above 
crown  2  feet.  A  construction  for  a  depth  of  keystone  of  3 
feet  showed,  for  the  loading  to  be  given,  that  the  line  of 
resistance  passed  outside  the  middle  third ;  hence,  for  a 
second  trial,  a  depth  of  arch  ring  of  3.5  feet  was  assumed. 
The  loading  omitted  the  pilot  wheel  and  consisted  of  the 
eight  drivers  on  the  left  half  of  the  arch,  viz.  :  15  tons  4.25 
feet  from  crown  ;  15  tons  10  feet  ;  15  tons  14.25  feet ;  and 
15  tons  19  feet,  all  to  left  of  crown  of  arch. 

The  line  of  resistance  nowhere  passed  outside  the 
middle  third  of  the  arch  ring.  At  the  springing  joints  the 
resultants  touch  the  lower  middle  third  limit  on  the 
loaded  side ;  and  the  upper  limit  on  the  unloaded  side ;  at 
the  crown  the  thrust  passes  through  the  centre  of  the 
crown  joint.  The  arch  thus  satisfies  all  the  conditions 
of  stability.  The  horizontal  component  of  thrust  at 
crown  =  279.3  cu.  ft.  =  19.55  tons,  and  th.3  vertical  com- 
ponent =  21  cu.  ft.  =  1.47  tons.  The  normal  component 
of  the  thrust  at  the  left  springing  =  413.5  cu.  ft.  =  28.95 
tons,  giving  an  intensity  at  the  intrados  of 


28  95 

2  — =16.5  tons, 

o.o 

which  is  within  the  proper  limit. 

This  arch  then  satisfies  all  conditions  for  this  loading. 

44.  Example  IV.     Stone  arch  of  100  ft.  span,  20  ft.  rise, 
radius  72.5  ft.,  depth  of  keystone  5  ft.,  and  height  of  sur- 
charge above  crown  2  ft.      The  Ir  adiug  consisted    of  one 
locomotive  and  tender,  as  specified  in  Art.  34,  covering  the 
left    half    of    arch,  the  right    half  being    unloaded.    The 
pilot  wheel  was  placed  8  feet  to  left  of  crown,  the  other 
wheels  following  in  order  at  the  distances  given  in  Art.  35. 
so  that  the  last  tender  wheel  was  barely  on  the  arch. 

The  line  of  resistance  was  ev<  rywhere  contained  within 
the  middle  third,  except  at  the  left  springing  joint  where 
it  passed  0.17  ft.  below  the  limit;  hence  to  satisfy  the 
middle  third  limit  the  arch  ring  should  be  increased  in 
depth,  say  3X  0.17=  .51ft.,  making  the  depth  5.5  ft. 

The  thrust  at  the  crown  joint  fell  0.4  ft.  below  centre 
for  the  arch  of  5  ft.  key,  its  horizontal  component  being 
689.7  cu.  ft.  =-•  48.279  tons,  and  the  vertical  component  26.5 
cu.  ft.  =1.855  ton.  The  intensity  of  thrust  at  lower  edge 
=  14.3  tons  per  sq.  foot. 

Tin?  thrust  at  the  left  springing  acted  1  foot  below  the 
centre  of  the  joint,  its  normal  component  being  1048  cu.  ft. 
=  73.36  tons.  It  acts  as  a  uniformly  increasing  stress 
over  a  depth  of  joint  =  3x1.5=4.5  ft.  ;  hence  the  aver- 
age stress  is  73. 36 -f- 4. 5=  16.3  and  the  intensity  at  lower 
edge  is  double  this,  or  32.6  tons  per  square  foot.  For  a  5.5 
ft.  key  the  intensity  is  much  less. 

45.  Example  V.     The  arch  of  Ex.  IV.  of  100  ft.  span,  20 
ft.  rise,  and  5  ft.   depth  of  key,  subjected  only  to  its  own 
weight. 

Here  we  need  consider  only  the  right  half  of  the  arch, 
since  the  thrust  at  the  crown,  from  consideration  of  sym- 
metry, must  be  horizontal  as  well  as  the  line  m  m'  (using 
the  designation  given  on  plate  for  another  span). 

Hence  for  an  assumed  horizontal  thrust  at  the  crown, 


178 

having  found  points  such  as  b  and  drawn  the  line  k  k'  i 
before,  we  determine  line  m  m'  for  points  6  exactly  as  \ 
found  kk'  for  points  a  ;  or  it  is  generally  shorter  to  add  up 
with  dividers  the  ordinates  to  points  b  above  a  trial  line  as 
kk',  and  subtract  from  the  sum  the  length  we  obtain  by 
adding  up  ordinates  to  points  b  below  k  k'.  The  difference 
divided  by  8  gives  the  amount  the  horizontal  m  m'  is  above 
or  below  k  k'  to  satisfy  the  condition  2  (,iub)  =  o.  The 
points  b  meant  above  are  b(),  b1(),  ....  blg.  As  a  test  the 
sum  of  the  ordinates  from  m  m'  to  points  b  above  should 
exactly  equal  the  sum  to  points  b  below  m  m'. 

The  sum  of  the  products  2  (ka  .  y)  is  made  out  exactly 
as  before.  Its  value  for  the  half  arch  in  this  case  is  302. 
Similarly  find  2  ^nib  .  y)  =  mb0  .  y()  -  mb]() .  yu)  +  mbn  .  yn 
f  mb^  .  y12  +  mbjg  .  ylg  -  (mbtt, '  yu  +  mbl6  .  Ju  mbM 
.  y  )  =  3S8,  the  ordinates  rub  being  measured  from  m  m'  up 
(-f )  or  down  (— )  to  points  6. 

On  diminishing  all  the  ordiuates  tub  in  the  ratio  of  Su±:( 
to  388  and  laying  them  off  from  k  k',  we  find  the  points  c 
through  which  the  resultants  on  the  joints  pass. 

The  point  of  thrust  at  the  crown  is  found  as  usual,  and 
the  new  pole  is  found  by  laying  off  on  a  horizontal  through 

3SS 
C'  a  distance  equal  to  old  pole  distance        ^— :.    The  points 

c  can  now  be  tested  and  the  resultants  produced  to  intersec- 
tion with  all  the  joints. 

It  is  interesting  to  compare  the  new  curve  of  the 
centres  of  pressure  for  the  bridge  unloaded,  with  that  found 
previously  for  the  load  on  the  left  half.  Thus  reiuembi  r- 
ing  that  the  depth  of  arch  ring  is  5  ft.,  one-sixth  of  which 
is  0.83  ft.  from  the  centre  to  curves  defining  middle  third 
limits,  the  following  tables  give  the  distance  measured  atony 
any  joint  from  its  centre  to  the  centre  of  pressure  of  the 
joint,  plus  distances  being  measured  upwards,  minus  dis- 
tances downwards.  The  upper  numbers,  under  the  joint 
numbers,  correspond  to  the  arch  unloaded  and  the  lower 
numbers  to  the  arch  loaded. 


171) 


KIGKT    SIDE. 


C  i  own 

9 

10 

11 

14 

i'» 

17 

—.27 

—  3  ' 

-.20 

() 

K20 

.  if» 

4-.  32 

(I 

_-,, 

—.4(1 

—  .57 

—.62 

-.15 

-f.05 

+.10 

.±! 

.i:. 

LEFT    SIDE. 


7 

G 

.j 

4 

3 

2 

1 

-.30 

—  .'20 
+.18 

II 
+.W) 

+.20 

+_75 

+.35 

0 

+  13 

—.22 

—.65 

-  .4(1 
1.00 

It  will  be  observed  at  joints  0,  5  and  10,  where  the  ceu 
hvs  of  pressure  for  bridge  loaded  are  farthest  from  centre 
of  joints,  that  when  the  live  load  comes  on  the  centres  of 
pressure  remain  on  the  same  side  of  the  centre  line  of  the 
arch  ring  as  for  bridge  unloaded.  At  many  other  joints  as 
17  the  reverse  obtains.  Hence,  if  the  arch  was  not  circular, 
but  of  such  a  figure  that,  for  bridge  unloaded,  its  centre 
line  would  be  the  focus  of  the  centres  of  pressure  on  the 
joints,  the  departure  of  the  line  of  resistance  for  bridge 
loaded  at  joints  0,  5  and  10  would  not  be  so  great  as  above, 
though  at  joint  17  (which  is  often  a  critical  joint),  and  at 
some  other  points,  it  would  beigreater.  Therefore  such  a 
design  would  often  permit  of  smaller  arch  rings,  such  that 
the  line  of  resistance  for  the  bridge  loaded  in  any  way  could 
still  be  inscribed  in  the  middle  third.  However,  in  some  of 
the  bridges  examined  (Exs.  II.  and  III.)  the  centre  of  pres- 
sure on  joint  17,  for  bridge  loaded,  passed  through  the  upper 
middle  third  limit,  so  that  if  an  arch  having  its  centre  line, 
the  line  of  resistance  for  arch  unloaded,  was  used  here  of 
the  same  depth  as  before,  the  centre  of  pressure  on  joint  17 
would  leave  the  middle  third,  and  the  arch  would  not  be  as 
stable  as  before. 

As  we  cannot  tell,  without  a  special  investigation  of 
this  kind,  which  design  will  prove  the  most  economical,  it 
is  well  to  hold  on  to  the  segmental  circular  arch  until  the 
others,  for  ail  kinds  of  loading,  are  proved  the  most  ecouom 


180 

ical,  particularly  as  it  is  much  easier  to  construct,  and  the 
economy,  if  any,  in  replacing  it  by  the  other,  must  be  small, 
Writers  generally,  in  advocating  the  catenarian  curves,  have 
not  properly  considered  the  preponderating  influence  of 
heavy  eccentric  loading. 

40.  An  examination  of  the  lines  of  re- 
sistance in  all  the  preceding  examples  fails 
to  indicate  any  simple  approximate  rule  for 
constructing  them  without  recurring  to  the 
theory  of  the  solid  arch.  I  have  stated 
elsewhere,  partly  on  the  strength  of  a  few 
constructions  after  the  theory  of  the  solid 
arch,  for  uniform  loads  or  comparatively 
light  eccentric  loads,  that  "  it  seems  highly 
probable  that  the  actual  line  of  resistance 
is  confined  within  such  limiting  curves,  ap- 
proximately equidistant  from  the  centre 
line  of  the  arch  ring  that  only  one  line  of 
resistance  can  be  drawn  therein."  From 
the  constructions  above  this  rule  is  found 
to  indicate  very  roughly  about  the  posi- 
tion, but  it  is  not  precise  enough  in  prac- 
tice. Thus  for  the  100  ft.  span  above  un- 
loaded, the  true  curve  passes  .4  below  the 
C2ntre  line  (measured,  not  vertically,  but 
along  the  joint)  at  the  springs,  .35  above  at 
joints  4  and  13  and  .37  below  at  the  crown ; 


181 

but  the  divergences  are  much  greater  at  the 
joints  of  rupture  (0,  4  or  o,  9  and  17)  for 
the  arch  heavily  loaded  on  one  side,  as  we 
see  from  the  table,  and  the  same  thing  is 
shown  on  the  plate  for  the  12.5  feet  span. 
Hence  we  cannot  state  precisely  that  if 
a  line  of  resistance  can  be  inscribed  in  the 
middle  third,  the  true  line  of  resistance 
will  be  found  in  the  middle  third.  It  is  in 
fact  plain  from  the  above  constructions 
that  if  only  one  line  of  resistance  can  be 
inscribed  in  the  middle  third,  the  true  line 
will  pass  outside  of  it  at  certain  points;  for 
the  first  line  touches  the  curves  limiting 
the  middle  third  at  all  the  joints  of  rup- 
ture, as  it  corresponds  to  both  the  maxi- 
mum and  minimum  of  the  thrust  within 
those  limits,  whereas  the  true  curve  does 
not  at  all  the  joints,  hence  it  cannot  agree 
with  the  former  and  hence  must  lie  outside 
the  middle  third  limits,  since  by  assump- 
tion only  one  line  of  resistance  can  be 
drawn  therein. 

We  can  appropriately  quote  here  Wink- 
ler's    notable   theorem  published  in  1879, 


which    is   practically   true   for    segmental 
arches  of  constant  cross-section.     The  the- 
ory is  given  in  full  in  an  article  by  Prof. 
Geo.  F.  Swain  on  the  stone  arch,  in  Van 
Nostrand's  Magazine  for  October,  18SO. 
Winkler's  theorem  is  as  follows: 
That  line  of  resistance  is  approximately 
the  true  one  which  lies  nearest  the  centre 
li)te  of  the  arch  ring  ax  deter  rained  by  the 
method  of  least  squares. 

This  remarkable  theorem  is  easily  dem- 
onstrated by  aid  of  the  theory  of  elasticity, 
and  while  it  is  no  aid  practically  in  pre- 
cisely fixing  the  true  line  of  resistance,  yet 
the  conclusions  are  valuable  as  confirming, 
in  a  general  way,  the  preceding  construc- 
tions. 

47.  The  method  of  finding  the  true  re- 
sistance line  given  in  this  chapter  is  per- 
fectly general  and  applies  to  any  form  of 
arch  of  constant  cross- section.  The  defect 
in  the  method  practically  is  that  the  most 
hurtful  position  of  the  live  load  cannot  be 
readily  ascertained.  It  is  true  that  for  sin- 
gle loads  the  method  of  this  chapter  will 
give  the  quantities  c,,  y  and  c2  as  defined 


in  Chap.  IV.,*  from  which  we  may  make 
out  a  table  and  proceed  as  in  the  preceding 
chapter.  This  method  is  very  long  and  is 
rarely  needed,  as  circular  arches  are  gener- 
ally built;  and  for  these  the  quantities  c1?  y 
and  c,  can  be  readily  found  from  existing 
tables  and  formulas. 

The  positions  of  live  loads  assumed  in 
this  chapter  simply  followed  the  roui^h 
rule  of  putting  the  heaviest  part  of  the 
load  over  the  middle  of  the  haunches.  The 
constructions  resulting  were  all  made  be- 
fore the  method  of  the  preceding  chapter 
was  developed.  It  is  gratifying  to  note 
thai  the  conclusions  as  to  depth  of  key, 
etc.,  are  almost  identical  with  those  of 
Chap.  IV.,  where  the  most  hurtful  position 
of  the  live  load  was  carefully  ascertained. 


*  S^H    such    constructions    in    "Theory  of    Solid    and 
Braet-d  Elastic  Arches, ''  p.  79. 


184 

APPENDIX. 

The  writer,  in  1874,  performed  some  experiments  on 
light  wooden  arches  at  the  limit  of  stability,  which  tend  to 
confirm  theory  and  are  instructive  in  many  ways.  They 
will  be  given  in  full  below.  The  experiments  were  made 
with  great  care ;  the  voussoirs  being  accurately  cut,  the 
span  kept  invariable  and  horizontal,  piers  vertical,  and 
the  weight  applied  very  gently  and  without  shock. 

To  avoid  mistake  the  following  momenclature  will  be 
adopted  : 

Depth  of  a  voussoir  is  the  dimension  in  the  direction  of 
the  radius  of  the  intrados  _[_  to  the  axis  of  the  arch. 

Thickness  of  an  arch  is  the  dimension  i!  to  the  axis  of 
the  arch. 

'  Width  of  a  pier  is  its  horizontal  dimension  J_  to  the  axis 
of  the  arch. 

Height  is  measured  vertically. 

The  dimensions  will  all  be  given  in  iuchi.  s. 

A  Gothic  arch  (Fig.  28)  of  14  in.  span,  and  12.12  in. 
rise,  was  cut  out  of  a  poplar  (tulip  tree)  plank,  3.65  in. 
thick,  consisting  of  8  voussoirs,  each  3. Go  thick,  i>  deep, 
and  4.U8  along  their  centre  line  from  middle  to  middle 
of  joint  :  each  vousscir  weighing  .52  Ib.  Quite  a 
number  of  voussoirs  were  cut  out  of  the  same  layers  of 
fibres  and  those  selected  that  weighed  exactly  the  same  : 
the  voussoir  to  be  tried  being  hung  to  one  end  of  a  delicate 
balance  beam,  with  a  voussoir  of  the  standard  weight  at  the 
other  end.  The  two  voussoirs  at  the  crown  not  being  cut 
out  of  the  same  layers  of  fibres  as  the  others,  were  shaved 
oft'  about  the  middle  of  the  extrados  (not  touching  the  joints) 
so  as  to  weigh  exactly  1  voussoir  of  the  standard  weight  and 
their  centres  of  gravity  were  found  experimentally,  and 
found  to  be  at  exactly  similar  points  in  both  voussoirs,  so 
that  the  entire  arch  was  symmetrical  as  to  the  crown. 

The  centres  of  gravity  of  the  other  voussoirs  are  taken 
on  the  arc  of  a  circle  passing  through  the  middle  of  the 
joints,  and  for  >my  voussoir,  equidistant  from  the  joints 
bounding  that  voussoir.  For  voussoirs  whose  sides  are 


185 

little  inclined  this  is  sufficiently  near  the  truth,  and  by 
dividing  the  arch  ring  into  a  sufficient  number  of  artificial 
roussoirs  the  result  may  be  made  as  accurate  as  we 
please.  Still  as  no  wood  is  homogeneous  the  results  can 
only  be  regarded  as  approximate  as  compared  with  the 
hypothetical  homogeneous  arch,  still  sufficiently  near  to 
establish  the  laws  heretofore  demonstrated. 

When  this  arch  was  set  up  the  joints  apparently 
fitted  perfectly,  and  on  placing  a  drawing-board  by  the 
side  of  the  arch  and  tracing  off  its  contour  curves,  it  was 
found  to  be  a  perfect  Gothic  whose  arcs,  composing  the 
contour  curves  were  correct  arcs  of  circles  described  from 
the  springing  points  opposite. 

A  number  of  rectangular  wooden  bricks  of  exactly  1 
voussoir  in  weight,  of  various  sizes,  were  also  cut  out,  as 
well  as  half  bricks,  quarter  bricks,  etc.,  and  some  solid 
rectangular  piers  of  various  dimensions. 

A  voussoir  is  taken  as  the  unit  of  weight. 

In  experiments  where  weights  were  placed  upon  the 
top  of  the  arch,  an  assistant  added  brick  after  brick, 
carefully  balancing  the  load  at  the  top  on  either  side  by 
the  fingers  until  the  arch  reached  its  balancing  point ; 
i.  c.,  the  point  where  it  stood  with  the  weight,  but  fell  with 
a  slight  jarring. 

The  two  bottom  voussoirs  were,  when  necessary,  kept 
from  sliding  by  two  fastening  tacks  being  driven  into  the 
board  on  which  the  ur.'li  rested,  pressing  against  the  arch 
.03  above  the  springing  line,  or  so  little  that  it  may  be  dis- 
regarded. The  board  was  carefully  levelled  at  every  exper- 
iment by  a  spirit  level,  and  the  span  kept  invariably  at 
14  in. 

There  was  little  or  no  -vibration  in  the  room. 

Ftrst  Experiment. — With  8.2  voussoirs  on  the  summit 
of  the  arch  it  stood,  though  fell  with  8.3  voussoirs  on  the 
summit  ;  rotating  on  joints  2  on  iutrado  al  edge,  join's  4 
at  the  extrados  and  at  the  upper  edge  of  the  crown  joint, 
the  arch  being  forced  out  at  the  haunches  and  falling  at 
the  crown.  (See  Fig.  28.) 


The  following  table  gives  in  its  first  column  the  number 
of  joint  from  the  crown;  columns,  the  elementary  weights 
(4.1  voussoir  being  the  weight  on  the  summit  that  goes  to 
each  abutment,  the  weight  of  each  voussoir  being  taken  as 
unity)  ;  column  in  gives  tho  horizontal  distance  from  the 
crown  to  the  centre  of  gravity  of  each  voussoir  with  its 
load,  if  any,  which,  in  this  case,  is  also  the  moment  in 
reference  to  the  vertical  through  the  crown  of  each  voussoir. 
Columns  8,  M,  and  C  have  been  before  « -xplained: 


s 

m 

s 

M 

1 

4.1 

0.00 
i  Tn 

4.1 
^  i 

0,00 

1   r/i 

•  ».  <  ih 

00 

1. 

1. 
1. 

1. 

8.1 


1.70 

4.68 
6.79 
7.88 


5.1 

G.I 
7.1 
8.1 


1.70 

6.38 
13.17 
21.05 


.33 
1.04 
1.85 


is; 

Try  a  line  of  resistance,  passing  0.1  from  the  upper 
edge  of  crown  joint  and  0.1  from  the  extrados  edge  of 
the  joint  at  the  springing.  It  is  found  to  cut  joint  2  at 
0.1  from  the  intrados. 

From  joint  0  to  joint  2  the  line  of  pressures  corres- 
ponds to  the  minimum  of  the  trust ;  from  joint  2  to  joint 
4,  to  the  maximum  within  limiting  curves  0.1  from  in- 
trados and  extrados  respectively.  (Art.  20.) 

Sliding  would  have  occurred  on  joint  4,  as  the  resultant 
on  that  joint  made  an  angle  of  21  deg.  with  the  normal, 
but  for  the  tacks  before  mentioned. 

The  diagrams  for  this  and  all  the  following  experiments 
were  drawn  to  a  scale  of  one-third  the  natural  size,  except 
in  the  case  of  some  of  the  pier  experiments. 

It  may  pertinently  be  "enquired,  why  at  the  limit  of 
stability,  the  centres  of  pressure  should  not  be  found  at 
the  very  edges  of  the  joints  in  place  of  being  ().".  inch 
from  those  edges  ?  The  answer  is  simple :  We  have  seen 
in  Art.  21  that  when  the  centre  of  pressure  on  a  joint  leaves 
the  middle  third,  the  joint  begins  to  open,  and  this  open- 
ing is  quite  perceptible  when  this  centre  of  pressure  is 
very  near  the  edge.  This  opening  of  the  joints  causes  a 
deformation  of  the  arch  ring,  so  that  the  figure  just  before 
rotation  occurred  is  not  that  assumed  in  the  drawing.  If 
the  deformation  had  been  known  at  the  instant  of  rupture,  so 
that  the  true  figure  could  have  been  drawn,  then  the  line  of 
resistance  would  have  passed  through  the  very  edges  of  the 
joints  0,  2  and  4,  as  they  alone  were  bearing  at  the  time. 
No  attempt  was  made  to  find  the  deformed  figure  ;  in  fact, 
it  varied  so  rapidly  just  before  rupture  that  it  would  have 
been  impossible  to  have  found  it.  Similar  remarks  and 
explanations  apply  to  all  the  subsequent  experiments. 

Second  Experiment. — With  the  two  voussoirs  at  the 
crown  in  one  solid  piece,  the  arch  could  not  give  by  rota- 
tion, as  the  lower  edge  of  crown  joint  could  not  open. 
With  a  sufficient  pressure  on  the  crown,  there  was  sliding 
along  joints  1,  the  coefficient  of  friction  being  small  for 
these  wooden  bl  cks. 


188 


We  evidently  have  here  a  sufficient  reason  for  making 
the  keystone  in  one  solid  piece. 

Third  Experiment. — On  placing  a  knife  edge  against  a 
notch  .03  deep,  cut  into  the  bottom  voussoir,  0.4  above  the 
springing  line,  on  each  side,  the  arch  balanced  with  11.1 
voussoirs  on  the  summit.  The  line  of  resistance  must  now 
pass  through  the  knife  edges,  and  it  will  be  found  on 
constructing  a  diagram  that  it  will  pass  about  0.1  fromedpcs 
at  joints  0  and  2,  as  before. 


Fig.  29 


Fourth  Experiment  — (Fig.  29.)  The  same  arch  stood, 
being  very  nearly  on  the  balancing  point,  on  solid  piers  In. 
high.  1.9  wide,  and  3.65  thick,  each  pier  weighing  2.3  vous- 
soirs, the  intrados  at  the  springing  being  at  the  inner  edge 
of  pier.  The  piers  were  made  vertical  by  a  spirit  level,  and 
their  tops  were  upon  the  same  level  in  every  experiment 
given. 


189 


in   the   following  table   the  pier  is  included  opposite 
joint  5  of  the  first  column  : 


s 

c 

m             S 

M 

C 

I 

I 

1 
1 

2.3 

1.7 

4.68 
6.79 

7.88 
7.95 

1.70           1 
4.f>8           2 
6.79           3 
7.88           4 
18.28           0.3 

1.70 
6.38 
13.17 
21.05 
39.33 

1.7 
3.19 
4.39 
5.26 
6.24 

6.3 

39.33 

A  line  of  resistance  0.1  from  edges  of  joints  o  and  8, 
cuts  the  base  of  the  pier  0.2  from  its  outer  edge. 

Fifth  Experiment.— With  piers  40.47  in.  high,  3.65  wide, 
and  1.9  thick,  weighing  10.1  voussoirs  each,  with  the  intra- 
dos  of  arch  at  springing  on  a  line  with  inner  edge  of  pier, 
the  same  arch  balanced.  The  pier  was  built  of  a  solid  block 
22  in.  high  and  5  bricks  placed  on  top,  one  above  the  other 
to  make  up  the  40.47  in  height. 

A  line  of  resistance  drawn  .1  from  summit  and  .1  from 
intrados  at  joint  3  passes  .5  from  outer  edge  of  pier,  or 
about  1-7  width  of  pier. 

If  the  figure  of  the  deformed  arch  could  have  been 
drawn  at  the  instant  of  rupture  the  line  would  have 
passed  through  the  very  edges. 

Sixth  Experiment.— The  pier  of  Exp.  4  (Fig.  29)  was 
moved  outward  (from  the  axis  of  the  arch)  >  o  that  when 
its  inner  edge  was  .1  from  the  springing,  it  stood  with  no 
weight  on  the  summit ;  when  it  was  .4  from  edge,  it  stood 
with  .5  vs.,  fell  with  .6  vs.  ;  .5  from  edge,  balanced  with  .7r> 
.vs.  :  .6  from  edge  balanced  with  .75  vs.  ;  .7  from  edge 
balanced  with  .37  vs.  ;  1.0  from  edge  balanced  with  .12  vs. 

On  constructing  the  table  and  diagram  as  above  for  the 
load  .75  vs.,  we  find  the  centre  of  pressure  on  joint  4,  .63 
from  the  inner  edge,  or  slightly  over  the  extreme  limit 
above,  as  should  be  the  case. 

Seventh  Experiment.— The  same  arch  stood  easily  with 
.75  vs.  on  the  summit,  on  solid  piers,  22.  high,  3.65  wide, 
and  1.9  thick,  each  weighing  5.1  vs. ;  the  arch  fell  with  the 
addition  of  .12  vs.  more. 


190 

On  constructing  this  figure  it  will  be  found  that  the 
line  of  centres  of  pressure,  assumed  0.1  from  edges  of 
joints  0  and  3  as  before,  passes  .63  from  inner  edge  of 
springing  jc  hit  (us  was  stated  above)  and  cuts  the  base  of 
pier  .39  from  its  outer  edge  or  about  1-9  the  width  of  pier. 

Enjhth  Ejrparhnent. — On  moving  this  pier  back  as  in  the 
(3th  KJ-I>,  : 

0.47  the  arch  balanc*  d  with  1.12  vs. 

0.53  "         "  "  "     1.25  " 

.5'.)   "          "  "  "      1.25   " 

.63  "         "  "  "     1.25  " 

.7     "         "  "  "     1.12  " 

1.        «         "  "     1.IM)  •* 

On  constructing  the  line  of  resistance  for  a  weight  of 
1.25  at  the  apex,  passing  0.1  from  the  edge  of  joints  0  and  3 
as  before,  it  will  be  found  that  the  centre  of  pressure  on 
joint  4  is  .7  from  the  edge,  again  slightly  over  the  extreme 
limit  .63  found  by  experiment. 

It  is  evident  from  an  inspection  of  the  arches  in  churches 
that  constructors  were  well  aware  that  a  higher  pier  might 
be  used  when  its  inner  edge  was  moved  back  a  certain  dis- 
tance from  the  springing,  which  is  equivalent  to  what  we 
have  established  above. 

Ninth  Experiment.—  With  the  pier  used  in  E-JCI>.  4.  and 
the  same  arch,  excepting  that  the  two  voussoirs  at  the 
crown  were  in  one  piece,  the  arch  and  pier  just  balanced  as 
in  Exp.  4.  In  fact  the  arch  and  pier  can  easily  rotate  on  the 
third  joint  and  the  outer  edge  of  pier. 

Tenth  Experiment. --The  same  arch  with  piers  1.98  wide, 
7.5  high  and  thickness  of  arch,  each  weighing  2  vs.,  stood 
easily  when  a  cylindrical  pin  .03  in  diameter  was  placed 
at  the  lower  edge  of  crown  joint.  This  joint  bore  at  no  other 
point,  hence  the  line  of  resistance  passes  through  the  pin. 
Assuming  it  to  pass  .1  from  the  edge  of  joint  3.  the  con- 
struction will  show  that  it  cuts  the  springing  joint  .6  from 
inner  edge  and  the  base  of  pier  .15  from  its  outer  edge. 

The  experiments    that  we    have    just  considered  very 


191 

clearly  indicate  the  fallacy  of  that  theory  which  supposes 
that  if  a  line  of  resistance  passes  outside  the  inntr  third  of 
the  arch  ring,  that  it  must  fall.  On  the  contrary,  in  every 
case  of  the  stability  of  the  arches  previously  given,  it  is 
impossible  to  draw  a  line  of  resistance  everywhere  contained 
within  the  inner  third  of  the  arch  ring. 

Eleventh  Experiment. — Fig.  oU.  With  this  same  Gothic 
arch  a  segniental  circular  arch  was  now  made  of  24.24  in. 
span  and  7  in.  rise  ;  the  voussoirs  being  as  before  2.  deep 
and  :>.G5  thick. 

With  7.C  vs.  on  the  summit,  this  arch  balanced;  the 
weight  being  placed  on  a  small  stick  resting  on  the  trammit. 
With  a  greater  weight  the  rotation  occurred  on  joints  0,  2 
and  4,  the  crown  falling. 


1 

2 
3 
4 

s. 

m. 

s. 

M. 

C. 

3.8 
1. 
1. 
1. 
1. 

0.00 
2.03 
5.90 

'.».  us 
12.11 

3.8 
4.8 

5.8 
(5.8 

7.s 

0.00 
2.03 
7  <>3 
17.31 
29.42 

.00 
.4-2 
137 
2.55 
3.77 

Fio.  30 


192 

On  trial  it  was  found  that  the  true  line  of  resistance 
passes  .15  from  the  edges  at  joints  0,  4  and  2  ;  giving  the 
characteristics  of  both  a  maximum  and  a  minimum  thrust. 

The  ends  of  this  arch  required  fastening  tacks  thrust 
into  the  board  and  pressing  against  voussoirs  4,  .03  above 
the  springing  as  in  the  first  exp.,  with  the  Gothic,  to  pre- 
vent sliding.  The  thrust  on  joint  4  made  an  angle  of  50  w 
with  the  normal  to  that  joint. 

Twelfth  Experiment. — With  this  arch  resting  on  piers 
3  63  wide,  5.8  high  and  2.  thick,  each  weighing  1.5  vs.,  the 
inner  edge  of  pier  being  on  a  line  with  the  springing,  the 
arch  balanced  with  .5  vs.  on  the  summit. 

We  find,  by  constructing  a  line  of  resistance  passing 
.15  from  summit  and  the  intrados  at  the  third  joint,  that 
it  cuts  the  base  of  pier  .24  from  its  outer  edge. 

Thirteenth  Experiment. — To  form  some  idea  of  the 
action  of  mortar  of  different  degrees  of  hardness,  pieces  of 
cloth  .07  thick  when  not  pressed,  and  .04  thick  when 
pressed  between  two  flat  surfaces  by  the  hands  were  put 
between  the  joints  of  the  Gothic  arch  (Fig.  28),  each  piece 
weighing  .015  voussoir. 

The  span  was  then  altered  until  the  joints  were  all 
close,  when  it  was  found  to  be  14.57,  the  rise  to  the  apex 
being  14.55.  On  placing  a  drawing-board  by  the  side  of 
this  arch  and  tracing  its  contour  curves,  they  were  found 
to  be  very  nearly  arcs  of  circles,  though  not  with  their 
centres  at  the  springing  points.  To  locate  them  ;  measure 
horizontally  from  the  springing  points  .32  towards  the 
middle  of  the  span,  and  then  vertically  downwards  0.1  to 
the  centres,  from  which  the  arch  may  be  drawn.  . 

This  arch  balanced  with  4.0  vs.  at  apex;  fell  with  4.05 
vs.  The  limiting  lines  to  the  curve  of  resistance  was  found 
to  be  distant  .3  =  1-7  depth  of  joint  from  the  contour 
curves,  at  its  nearest  approach  to  them. 

This  arch  spread  outwards  upon  the  application  of  the 
weights,  joint  2  being  the  point  of  rupture  at  the  haunches  ; 
hence  it  is  evident  that  if  there  had  been  a  solid  spandrel, 
or  in  this  case,  simply  the  pressure  of  the  hands,  to  resist 


193 

this  spreading,  that  the  arch  would  not  have  fallen.  The 
spandrel  would  have  supplied  horizontal  forces  in  addition 
to  the  vertical^ones  due  to  its  weight. 

If  the  spandrel  were  not  solidly  built,  at  least  up  to 
joint  '2,  there  would  necessarily  be  derangement  of  the 
arch. 

The  curves  of  resistance  were  drawn  in  all  the  fore- 
going experiments,  not  taking  into  consideration  the  last 
mentioned  derangement  of  the  arch,  which  would  have 
caused  this  curve  to  pass  nearer  the  edges  or  exactly 
through  them. 

In  fact,  in  most  of  the  experiments,  just  before  rotating, 
the  edges  alone  seemed  to  be  bearing.  In  the  case  of  the 
simple  Gothic,  without  cloth  joints,  when  a  sufficient 
weight  was  applied  at  the  summit,  the  joint  there  and 
joint  2  opened  sensibly  before  the  balancing  weight  was 
put  on.  The  segmeiital  arch  flew  out  at  the  second  joints, 
falling  at  the  crown,  only  opening  when  near  the  balancing 
point. 

Isolated  weights  applied  at  the  summit  do  not  occur 
in  practice,  and  it  is  hardly  probable  that  a  well-built 
viaduct,  whose  intrados  is  a  segment  of  a  circle  with  thin 
joints,  will  spread  appreciably  after  the  mortar  has  well 
set  ;  and  this  is  necessarily  a  stronger  form  of  arch  than 
the  semi-circular,  elliptical,  or  hydrostatic,  where;  the 
spandrel  thrust  is  generally  required  to  cause  stability. 

If  the  latter  profiles  are  desired,  let  the  depth  of  the 
voussoirs  be  increased  towards  the  abutment,  so  as  to 
keep  the  line  of  resistance  within  the  proper  limits  of  the 
arch  ring,  when  the  constructor  will  be  assured  of  stability. 

It  certainly  seems  singular,  that  engineers  should  ever 
rci-niti'iHdul  an  aivh  like  the  hydrostatic,  which  necessarily 
ri -quires  a  very  effective  spandrel  thrust  to  keep  the  arch 
from  tumbling  down. 

The  spandrels  must  in  such  cases  be  built  with  the  same 
'•are  used  with  the  arch  stones,  thus  increasing  the  expense, 
while  really  losing  in  stnngth. 

Fourteenth    Experiment. — In    the  joints    of    the    same 


194 

Gothic*  arch,  pieces  of  soft  woolen  cloth  .15  thick  when 
not  pressed,  and  .1  when  pressed  hard  between  two  bri  -ks 
by  the  hands,  were  next  inserted,  each  piece  of  cl:>tli 
weighing  .027  voussoir.  The  span,  when  the  joints  w«  r> 
close,  was  found  to  be  15  in.  ;  rise  to  apex,  14.63,  The 
centres  for  describing  the  contour  curves  were  1.07  in.  from 
the  springing  points  measured  horizontally  towards  the 
middle  of  span. 

This  arch  balanced  with  2.3  vs.  on  the  apex. 

Assuming  this  arch  to  preserve  its  figure,  the  curve  of 
resistance  passes  about  one-fourth  of  the  depth  of  joint 
from  the  edges  at  its  nearest  approach  to  them. 

This  experiment  gives  us  some  idea  of  the  effect  of 
thick  plastic  mortar  joints  or  of  uncentreing  an  arch  with 
fresh  mortar  joints. 

Fifteenth  Experiment. — A  Gothic  arch  of  about  half  the 
dimensions  of  the  first  given  in  Exp.  1  was  cut  out,  really 
lief  ore  the  arch  we  have  just  been  considering. 

It  was  not  found  to  be  symmetrical  as  to  weight,  one- 
half  weighing  1-32  of  the  whole  arch  more  then  the.  other 
half.  Still  as  arches  in  practice  are  unsymnu -tri'-al  as  to 
weight  at  least  it  will  be  interesting  to  know,  that  assum- 
ing this  arch  to  be  symmetrical,  the  curve  of  pressures 
passes  .075  from  the  edges  at  joints  of  rupture,  mora- 
lly with  weights  at  the  apex. 

All  the  preceding  experiments  were  repeated  with  t;:is 
arch  and  the  same  laws  approximate^  established. 

In  the  experiment  with  the  cloth  joints  the  cloth  was 
.05  thick  not  pressed  ;  .04  when  pressed  hard  b 
The  curve  of  resistance  was  found  to  pass  .1  from  tne 
edges  at  the  joints  of  rupture,  with  a  weight  on  the  ,-i>ex, 
and  nearly  so  in  the  pier  experiment  with  no  weight  on  the 
apex. 

Sixteenth  Experiment.—  The  Gothic  arch  given  by  Fig. 
28  will  now  be  considered  with  an  uusymmetrieal  loud.  A 
stout  needle  was  thrust  into  the  second  voussoir  from  the 
crown  on  the  right  side,  in  the,  direction  of  a  vertical 
through  its  centre  of  gravity,  as  represent*  d  in  Fit:.  31.  With 


L95 

a  weight  of  3.3  vs.  on  the  top  of  the  needle,  the  arch 
balanct-d,  opening  at  summit  and  lower  end  of  joint  1  on 
the  right.  The  voussoir  to  which  the  weight  was  added 
would  have  slid  if  urns  had  not  been  thrust  into  the  edges 
of  its  joints,  thus  supplying  n  force  analogous  to  friction, 
though  not  interfering  at  all  with  rotation. 

We   now  form  the   following   tabl  s :    the    first    being 
condensed  from  the  one  referring  to  Lxp.  :;. 


LKFT    .-IDE. 


S 

C 

1.70 
3.19 
4.39 
5.26 

1 

•2 
3 
4 

1 
2 

3 
4 

196 


RIGHT    BIDE. 


s 

C 

M 

8 

M 

1 

2 
3 

4 

1. 

4.3 
1. 
1. 

1.7 

4,68 
6,79 

7.88 

1.70 
20.12 
6.79 

7.88 

1. 
5.3 
6.3 
7.3 

1.70 
21.82 
28.61 
36.49 

1.70 
4.12 
4.54 
5. 

7.3 


36.49 


A  line  of  resistance  can  be  drawn,  as  shown  in  the 
figure,  passing  .18  from  the  extrados  at  joints  4  on  left,  and 
1  on  right  and  .18  from  the  intrados  at  joints  0  and  3  on 
the  right. 

The  lower  edge  of  the  crown  joint  was  imperfect, 
being  the  only  imperfect  edge  in  the  arch,  and  this  may 
account  for  the  line  of  resistance  retreating  farther  in 
the  arch  than  for  a  load  in  the  summit  as  before  r.>n- 
sidered. 

The  thrust  on  joint  1,  on  the  right,  was  inclined  at 
an  angle  of  15°  to  the  normal  to  that  joint,  which  accounts 
for  the  sliding,  as  the  joints  were  planed  and  across  the 
grain. 

Seventeenth  E.rperimciii. — The  segmented  arch.  Fig.  :-:o, 
was  next  tried  with  the  eccentric  load. 

A  short  needle  was  thrust  in  voussoir  2  on  the  left,  in 
the  direction  of  the  vertical  through  its  centre  of  gravity,  as 
shown  in  Fig.  32 ;  the  arch  balanced  with  5.4  voussoirs  on 
the  top  of  this  needle. 

We  form  the  following  tables  : 


UIGHT  SIDE. 


s 

m 

S 

M         ,       C 

.1 

2 
3 
4 

1 
1 
1 
1 

2.03 
5.90 
9.38 
12.11 

1 
2 
3 
4 

2.03 
7.93 
17.31 
'J9.42 

2.03 

3.96 
5.77 
1     7.37 

198 


LEFT   SIDE. 


S 

M                   S 

M                C 

2 
3 
4 

1. 

6.4 
1. 
1. 

2.03 
37.76 
9.38 
12.11 

1. 
7.4 
8.4 
9.4 

2.  03 
39.79 
49.17 
61.28 

2.03 
5.38 
5.85 
6.52 

9.4  61,28 

The  voussoir  on  which  the  weight  was  placed  would 
have  slid  along  its  joints  but  for  pins  being  thrust  into  its 
edges  in  a  manner  that  did  not  interfere  with  rotation. 

A  line  of  resistance  was  drawn  that  passes  .15  from  the 
iutrados  at  joint  2  on  the  right  and  .2  distant  from  the 
edges  at  joints  4,  1  and  4;  hence  the  true  curve  will  prob- 
ably puss  about  .18  from  these  edges.  This  is  nearly  (.03 
difference)  what  we  obtained,  for  the  limits  from  the 
of  the  line  of  resistance  in  the  llth  Exp.,  Fig.  '3(\.  The 
thrust  on  joint  1  on  the  left  is  inclined  16°  .5  to  the  normal 
to  the  joint,  nearly  wh<vi:  we  found  before.  The  sliding  in 
this  and  the  last  experiment  only  occurred  just  before  the 
balancing  weight  was  applied  ;  the  line  of  pressures  travel- 
ling down  the  crown  joint  as  the  weight  was  increased,  un- 
til finally  the  direction  of  the  pressure  on  joint  1  exceeded 
the  complement  of  the  angle  of  friction. 

Eighteenth  Experiment.—  Figure  33  represents  two  rat- 
ters 9.92  in  length,  1.9  width  (dimension  in  plane  of  paper) 
and  3.60  thick,  leaning  against  each  other  at  the  top  and 
against  piers  7.7  high,  1.98  wide  and  3.(>  thick  at  their  bot- 
tom edge,  which  is  moved  back  0.6  from  the  edge  of  the  pier. 
The  horizontal  distance  between  the  vertical  piers  is  in  in., 
so  that  the  feet  of  the  rafters  are  11.2  apart.  Each  rafter 
weighed  2.3  vs.  ;  each  pier  2.  vs.  The  rafters  and  piers  just 
balanced  in  this  position. 

Reasoning  as  in  Art.  2,  we  see  that  the  thrust  at  the  upper 
edges  of  contact  of  the  rafters  is  horizontal ;  hence  draw  a 
vertical  line  through  the  centre  of  gravity  of  the  rafter 
equal  to  its  weight ;  the  resultant  on  the  lower  edge  of  the 
rafter  passes  through  this  edge,  and  combined  with  the 


199 


Fig.  33 


weight  of  the  pier  acting  through  its  centre  of  gravity,  gives 
the  resultant  thrust  on  the  base  of  the  pier.     In  thh- 
it  strikes  twenty-two  hundredths  (.'22)  from  its  outer  edire. 

This  experiment  was  performed  to  ascertain  whether 
the  resultant  on  the  ba-e  could  ever  be  drawn  through  the 
outer  edge  of  base  for  the  original  figure.  It  seemed  prob- 
able, as  the  centres  of  pressure  at  the  apex  and  top  of  the 
pier  were  absolutely  fixed,  and  there  was  only  one  real 


200 


joint  at  the  base  of  the  pier ;  but  we  see,  even  from  this 
case,  that  the  joint  opened  sufficiently  to  deform  the  origi- 
nal figure,  so  that  the  resultant  cannot  be  drawn  exactly 
through  the  outer  edge  for  the  original  figure.  This  should 
offer  a  valuable  hint  to  experimenters  and  constructors,  not 


201 

to  look  for  the  stability  in  similar  structures  that  the  the- 
ory of  ' '  rigid ' '  or  incompressible  bodies  would  give,  espec- 
ially structures  composed  of  a  great  number  of  blocks  with- 
out cementing  material. 

Nineteenth  Experiment. — Fig.  (34),  represents  a  r-fter  and 
pier  of  the  preceding  experiment ;  the  rafter  leaning  against 
a  vertical  rough  plastered  wall  by  its  edge,  the  lower  edge 
resting  on  the  pier  1.03  back  from  its  inner  edge.  This  was 
the  balancing  position. 

After  several  trials,  assuming  as  we  found  in  the  pre- 
ceding experiment,  that  the  resultant  strikes  .22  from  the 
outer  edge  of  the  base  of  pier,  it  was  found  that  the  direc- 
tion of  the  thrust  against  the  wal?  was  inclined  about  35°  to 
the  horizontal,  which  is  about  what  we  should  imagine  the 
angle  of  friction  of  the  edge  on  the  wall  to  be.  If  the  thrust 
at  the  upper  edge  be  assumed  horizontal  as  is  usual,  it  will 
be  found  that  the  final  resultant  passes  outside  the  base  of 
pier;  hence,  such  an  assumption  is  false.  The  construc- 
tion (Fig.  34),  will  also  show  that  .32  v.  of  the  rafter  is  sus- 
tained by  the  wall,  1.98  v.  being  supported  by  the  pier  :  i.  c. 
about  one  seventh  of 'the  weight  of  the  rafter  is  upheld  by 
the  friction  of  the  plastered  wall. 

On  leaning  a  half  arch  against  a  wall,  it  was  found  to 
balance  on  higher  piers  than  when  the  other  half  was  placed 
against  it, 


THIS  BOOK  IS  DUB  ON  THE  LAST  DATE 
STAMPED  BELOW 


AN  INITIAL  PINE  OF  25  CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.OO  ON  THE  SEVENTH  DAY 
OVERDUE. 


MAY  20  1940 


MAR  Z  -  1990 


*,'•>•••  y  ffffl 


\ 


jflL  APR  Ob 


YA  06822 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 


